Probability Basic Binomial Poisson and Continuous.ipynb

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   "source": [
    "##"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Simple, Joint, Marginal, Conditional probability and Bayes Theorem"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Simple Probability refers to the probability of occurrence of a simple event"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Probability of an event X, P(X) is given by\n",
    "$P(X) = \\frac{Number \\quad of \\quad observations \\quad in \\quad favor \\quad of \\quad an \\quad event \\quad X}{Total \\quad Number \\quad of \\quad observations}$ "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The data collected by an Advertisement agency has revealed that out of 2800 visitors, 56 visitors clicked on 1 Advertisement, 30 clicked on 2 advertisements and 14 clicked on 3 advertisements and the remaining did not click on any advertisement. \n",
    "\n",
    "Calculate\n",
    "* a) The probability that a visitor to the website will not click on any  advertisement.\n",
    "* b) The probability that a visitor to the website will click on an advertisement.\n",
    "* c) The probability that a visitor to the website will click on more than one advertisement."
   ]
  },
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   "metadata": {},
   "source": [
    "### Solution\n",
    "\n",
    "a) The probability that a visitor to the website will not click on any advertisement.\n",
    "* Here 56 visitors click on 1 advertisement\n",
    "* Here 30 visitors click on 2 advertisements\n",
    "* Here 14 visitors click on 3 advertisements\n",
    "* Hence, the total number of customers clicking on any advertisement is 100"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
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     "text": [
      "a. The probability that a visitor to the website will not click on any advertisement is 0.9643\n"
     ]
    }
   ],
   "source": [
    "FE1 = 2800 - 100\n",
    "TE  = 2800\n",
    "PE1 = round(FE1 / TE,4)\n",
    "print('a. The probability that a visitor to the website will not click on any advertisement is %1.4f' % PE1)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "b) The probability that a visitor to the website will click on an advertisement."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
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     "text": [
      "b. The probability that a visitor to the website will click on an advertisement is 0.0357\n"
     ]
    }
   ],
   "source": [
    "FE2 = 100 # 56 + 30 + 14\n",
    "TE  = 2800\n",
    "PE2 = round(FE2 / TE,4)\n",
    "print('b. The probability that a visitor to the website will click on an advertisement is %1.4f' % PE2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "c) The probability that a visitor to the website will click on more than one advertisement."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
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     "text": [
      "c. The probability that a visitor to the website will click on more than one advertisement. is 0.0157\n"
     ]
    }
   ],
   "source": [
    "FE3 = 44 # = 30 + 14\n",
    "TE  = 2800\n",
    "PE3 = round(FE3 / TE,4) \n",
    "print('c. The probability that a visitor to the website will click on more than one advertisement. is %1.4f' % PE3)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Joint Probability refers to the probability of occurrence involving two or more events"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Let A and B be the two events in a sample space. Then the joint probability if the two events denoted by P(A $\\cap$ B), is given by \n",
    "$P(A \\cap B) = \\frac{Number \\quad of \\quad observations \\quad in \\quad A \\cap B } {Total \\quad Number \\quad of \\quad observations}$ "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "At a popular company service center, a total of 100 complaints were received. 80 customers complained about late delivery of the items and 60 complained about poor product quality.\n",
    "\n",
    "* a) Calculate the probability that a customer complaint will be about both product quality and late delivery.\n",
    "* b) What is the probability that a complaint will be only about  late delivery?"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Solution:\n",
    "\n",
    "a) Calculate the probability that a customer complaint will be about both product quality and late delivery"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Let\n",
    "*    L    = Late delivery\n",
    "*    Q    = Poor quality\n",
    "*    n(L) = Number of cases in favour of L =  80\n",
    "*    n(Q) = Number of cases in favour of Q =  60\n",
    "*    N    = Total Number of complaints     = 100    "
   ]
  },
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   "source": [
    "$n(L \\cap Q)$ = (80 + 60) - 100 = 40 "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Probability that a customer complaint will be about both product quality and late delivery = $P(L \\cap Q)$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    " $P(L \\cap Q) =  \\frac{n(L \\cap Q)} {Total \\quad Number \\quad of \\quad observations}$ "
   ]
  },
  {
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     "text": [
      "a.Probability that a customer complaint will\n",
      "       be about both product quality and late delivery. is 0.4000\n"
     ]
    }
   ],
   "source": [
    "FE4 = 40 # = (80 + 60) - 100 \n",
    "TE  = 100\n",
    "PE4 = round(FE4 / TE,4) \n",
    "print('a.Probability that a customer complaint will\\n\\\n",
    "       be about both product quality and late delivery. is %1.4f' % PE4)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "b. What is the probability that a complaint will be only about late delivery"
   ]
  },
  {
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   "metadata": {
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     "text": [
      "b.probability that a complaint will be \n",
      "         only about late delivery. is 0.4000\n"
     ]
    }
   ],
   "source": [
    "# FE5 is the complaints about poor quality\n",
    "FE5 = 60\n",
    "TE  = 100\n",
    "PE5 = round(FE5 / TE,4) \n",
    "PE6 = 1 - PE5\n",
    "# So,1 - PE5 is the Probability that a customer complaint will be about\n",
    "#        both product quality and late delivery \n",
    "print('b.probability that a complaint will be \\n\\\n",
    "         only about late delivery. is %1.4f' % PE6)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "|Planned to purchase Apple iPhone Xs Max | Actually placed an order for Apple iPhone Xs Max- Yes |  Actually placed an order for Apple iPhone Xs Max - Yes | Total |\n",
    "| ------------- | ------------ | ---------- | -----|\n",
    "| Yes | 400 | 100 | 500 |\n",
    "| No | 200 | 1300 | 1500 |\n",
    "| Total | 600 | 1400 | 2000 |\n",
    "\n",
    "Calculate the joint probability of the people who planned to purchase and actually placed an order.\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "You observe from the above table, that 400 people planned to purchase and actually placed an order for Apple iPhone Xs Max is 400 out of 2000 people."
   ]
  },
  {
   "cell_type": "code",
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     "text": [
      "Joint probability of the people who planned to purchase and actually placed an order. is 0.2000\n"
     ]
    }
   ],
   "source": [
    "# FE6 is the number of people who planned and actually placed an order for Apple iPhone Xs Max\n",
    "FE6 = 400\n",
    "TE  = 2000\n",
    "PE7 = round(FE6 / TE,4) \n",
    "print('Joint probability of the people who planned to purchase and actually placed an order. is %1.4f' % PE7)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Marginal probability refers to the probability of an event without any condition"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "P(A) = P(A and $B_{1}$) + P(A and $B_{2}$) + P(A and $B_{3}$) + ... + P(A and $B_{k}$) \n",
    "where $B_{1}$, $B_{2}$, $B_{3}$, ..., $B_{k}$ are k mutually exclusive and collectively exhaustive events, defined as follows:\n",
    "\n",
    "* Two events are mutually exclusive if both the events cannot occur simultaneously.\n",
    "* A set of events are collectively exhaustive if one of the events must occur."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 3"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Use the purchase of **Apple iPhone Xs Max** table.\n",
    "What is the probability of planned to purchase **Apple iPhone Xs Max**?"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "P(planned to purchase Apple iPhone Xs Max) \n",
    " =   P(Planned to purchase Apple iPhone Xs Max and placed an order) + \n",
    "     P(Planned to purchase Apple iPhone Xs Max and not placed an order) "
   ]
  },
  {
   "cell_type": "code",
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   "outputs": [
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     "text": [
      "Marginal probability of the people who planned to purchase is 0.2500\n"
     ]
    }
   ],
   "source": [
    "# Let P  = P(planned to purchase Apple iPhone Xs Max)\n",
    "#     P1 = P(Planned to purchase Apple iPhone Xs Max and placed an order) \n",
    "#     P2 = P(Planned to purchase Apple iPhone Xs Max and not placed an order) \n",
    "P1 = 400 / 2000 \n",
    "P2 = 100 / 2000\n",
    "P  = P1 + P2\n",
    "print('Marginal probability of the people who planned to purchase is %1.4f' % P)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Note that you get the same result by adding the number of outcomes that make up the simple event *planned to purchase* and calculate the probability of that *simple event*."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### In class LAB : Practice Exercise 1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Consider the following Contingency table:**\n",
    "\n",
    "| Event | Event B | Event $B'$ |\n",
    "| ------- | ----- | ------ |\n",
    "| A | 10 | 20 |\n",
    "| $A'$ | 20 | 40 |"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "* a. What is the probability of A | B?\n",
    "* b. What is the probability of A | $B'$?\n",
    "* c. What is the probability of $A'$ | $B'$?\n",
    "* d. Are events A and B independent?"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Hint: Events A and B are independent, when P(A|B) = P(A)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### General addition rule"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "To get the probability of the event *A* or *B*, you need to consider the occurrence of either event A or B or both A and B."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### General addition rule"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "P(A or B)   = P(A) + P(B) - P(A and B)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "From set theory\n",
    "* P(A$\\cup$B) is the event that either A or B or both occur.\n",
    "* P(A$\\cap$B) is the event that both A and B occur at the same time.\n",
    "* Events A and B are mutually exclusive if they cannot happen at the same time:\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "P(A$\\cup$B) = P(A) + P(B) - P(A $\\cap$ B)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 4"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Use the purchase of **Apple iPhone Xs Max** table.\n",
    "What is the probability of planned to purchase **Apple iPhone Xs Max** or placed an order?"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "P(Planned to purchase Apple iPhone Xs Max or placed an order) <br/>\n",
    "= P(Placed an order) + P(Planned to purchase Apple iPhone Xs Max) -<br/>\n",
    "  P(Planned to purchase Apple iPhone Xs Max and placed an order)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Planned to purchase Apple iPhone Xs Max or placed an order is 0.3500\n"
     ]
    }
   ],
   "source": [
    "P1 =  500 / 2000 # P1 = P(Planned to purchase)\n",
    "P2 =  600 / 2000 # P2 = P(Placed an order)\n",
    "P3 =  400 / 2000 # P3 = P(Planned to purchase Apple iPhone Xs Max and placed an order)\n",
    "\n",
    "P  =  P1 + P2 - P3 # P = P(Planned to purchase Apple iPhone Xs Max or placed an order) \n",
    "print('Planned to purchase Apple iPhone Xs Max or placed an order is %1.4f' % P)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Conditional Probability refers to the probability of event A, given information about the occurrence of another event B"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Probability of A given B is written as P(A | B).\n",
    "\n",
    "$P(A\\mid B) = \\frac{P(A \\quad and \\quad B)}{P(B)}$\n",
    "\n",
    "where P(A and B) = Joint probability of A and B\n",
    "*     P(A)       = Marginal probability of A\n",
    "*     P(B)       = Marginal probability of B"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 5"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Use the purchase of **Apple iPhone Xs Max** table.\n",
    "\n",
    "Find the joint probability of the people who planned to purchase and actually placed an order, given that people planned to purchase."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "P(Actually placed an order | Planned to purchase) is 0.8000\n"
     ]
    }
   ],
   "source": [
    "# P1 = P(Actually placed an order | Planned to purchase)\n",
    "# P2 = Planned to purchase and actually placed an order\n",
    "# P3 = Planned to purchase\n",
    "P1   = (400 / 2000 )/ (500 / 2000)\n",
    "\n",
    "print('P(Actually placed an order | Planned to purchase) is %1.4f' % P1)     "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 6\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The following table describes loan default status at a bank and their marital status. \n",
    "\n",
    "| Marital Status | Loan Defaulted | Loan No Default | Marginal Total |\n",
    "| ----------- | ------ | ------- | -------- |\n",
    "| Single | 42 | 258 | 300 |\n",
    "| Married | 60 | 590 | 650 |\n",
    "| Divorced | 13 | 37 | 50 |\n",
    "| Marginal Total | 115 | 885 | 1000 |\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Based on the above table, calculate the probability of default  given divorced."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "P(Default | Divorced)  is 0.2600\n"
     ]
    }
   ],
   "source": [
    "# P1 = P(Default and Divorced)\n",
    "P1   = 13 / 50\n",
    "# P2 = P(Divorced)\n",
    "P2   = 50 / 1000\n",
    "#P3  = P(Default | Divorced) = P(Default and Divorced) / P(Divorced)\n",
    "P3   = (13/1000) / (50/1000)\n",
    "print('P(Default | Divorced)  is %1.4f' % P3)     "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Independent Events"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Two events, A and B are independent if and only if\n",
    "P(A | B) = P(A), \n",
    "\n",
    "where \n",
    "* P(A|B) is the conditional probability of A given B\n",
    "* P(A)   is the marginal probability of A"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Example: A student getting A grade in both Final Stats exam and in final Marketing exam"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 7\n",
    "\n",
    "What is the probability of getting a \"6\" in two consecutive trials when rolling a dice?\n",
    "\n",
    "For each roll of a dice:\n",
    "* Favorable events = {\"6\"}\n",
    "* Total number of outcomes = {\"1\",\"2\",\"3\",\"4\",\"5\",\"6\"}\n",
    "* Let P1 be the probability of getting a \"6\" in the first roll of dice.\n",
    "* Let P2 be the probability of getting a \"6\" in the second roll of dice.\n",
    "* Since first roll of dice does not influence the second roll of dice, these events are independent."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Getting a 6 in two consecutive rolls of dice is 0.0278\n"
     ]
    }
   ],
   "source": [
    "P1 = 1 / 6\n",
    "P2 = 1 / 6\n",
    "P   =  P1 * P2 # P = P(Getting a 6 in two consecutive rolls of dice)\n",
    "print('Getting a 6 in two consecutive rolls of dice is %1.4f' % P) "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### In class LAB :Practice Exercise 2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "** What is the probability of getting a 2 on the face of three dices  when they are rolled?\n",
    "Hint: A dice has six faces and contains values 1,2,3,4,5,6**"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### In class LAB :Practice Exercise 3"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "** You throw a die three times, what is the probability that one or more of your throws will come up with a 1?\n",
    "Hint: You need to calculate the probability of getting a 1 on at least one of the throws.**"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### In class LAB :Practice Exercise 4"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "** The following table describes loan default status at a Financial Institution and their Occupation status.\n",
    "Calculate the Ocupation status that has maximum joint probability of default.**\n",
    "\n",
    "| Occupation Status | Loan defaulted | Loan non-default | Total |\n",
    "| ----------------- | ---- | ----- | ---- |\n",
    "| Self Employed     | 80 | 240 | 320 |\n",
    "| Employed in Private Sector | 120 | 860 | 980 |\n",
    "| Employed in Government Sector | 200 | 3000 | 3200 |\n",
    "| Total | 400 | 4100 | 4500 | "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### In class LAB :Practice Exercise 5"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "** In the above contingency table, what is the conditional probability that**\n",
    "* a. What is the probability of Loan defaulted?\n",
    "* b. What is the conditional probability of default, given the occupation category **Self Employed**?"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Application of Probability Rules"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Association Rule Mining"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "By using simple probability concepts such as Joint probability and Conditional probability, we solve Market Basket Analysis and Recommender Systems. The algorithm used is **Association Rule Mining**. "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Market Basket Analysis** is used frequently by retailers to predict products a customer is likely to buy together to improve their sales.\n",
    "\n",
    "For example, if a customer buys **bread**, he/she is likely to buy **jam or butter**.\n",
    "\n",
    "**Recommender Systems** are models that produce list of recommendations to a customer on products such as moives, electronic items, etc.\n",
    "\n",
    "As you are aware, companies like **Amazon** benefitted significantly by using recommender systems."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Association Rule Mining"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "* This is a method of finding association between different entities in data.\n",
    "* In a retail context, this is a method of finding association relationships that exists in *items frequently purchased* or *frequently purchased items*.\n",
    "* Association Rule is a relationship of the form X -> Y (X implies Y), where X and Y are mutually exclusive SKU sets."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Creating Association Rules"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Point of Sale (PoS) data"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "| Invoice No | Milk | Bread | Butter | Jam |\n",
    "| --- | ---- | ---- | ---- | ----- |\n",
    "| 1 | Y | Y | N | N|\n",
    "| 2 | N | Y | Y | N|\n",
    "| 3 | N | Y | Y | N|\n",
    "| 4 | N | Y | N | Y|\n",
    "| 5 | Y | Y | N | N|\n",
    "| 6 | Y | Y | N | N|\n",
    "| 7 | N | Y | Y | N|\n",
    "| 8 | N | Y | Y | N|\n",
    "| 9 | N | Y | N | Y|\n",
    "| 10 | Y | N | N | N|"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "In the above table, Milk, Bread, Butter and Jam are the different SKUs sold by the store. Y means the item is purchased and N means the item is not purchased. \n",
    "The strength of the association between two mutually exclusive subsets can be measured using *Support*, *Confidence* and *Lift*."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Support between two sets is the joint probability of those events"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "* Support is the proportion of times, items A and B are purchased together.\n",
    "* Support(A, B) = P(A $\\cap$ B) = n(A $\\cap$ B) / N\n",
    "where n(A $\\cap$ B) is the number of times both item A and item B is purchased together."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 8\n",
    "From the above table, assume A to be Bread and B to be Milk.\n",
    "\n",
    "Support(A, B) = P(A ∩ B) "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Support(Bread and Milk)  is 0.3000\n"
     ]
    }
   ],
   "source": [
    "# Let S_A_B = P(A ∩ B)\n",
    "S_A_B = 3 / 10 #  Refer invoices 1,5,6\n",
    "print('Support(Bread and Milk)  is %1.4f' % S_A_B) "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Confidence is the conditional probability of purchasing item B given the item B is purchased."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Confidence(A, B) = P(A $\\cap$ B) / P(A) "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Assume C_A_B be the confidence of purchasing a item B\n",
      "\tgiven the item A is purchased.\n",
      "S_A_B =  P(A ∩ B) as discussed earlier.\n",
      "Confidence(Bread a∩d Milk)  is 0.3333\n"
     ]
    }
   ],
   "source": [
    "print('Assume C_A_B be the confidence of purchasing a item B') \n",
    "print('\\tgiven the item A is purchased.')\n",
    "print('S_A_B =  P(A ∩ B) as discussed earlier.')\n",
    "P1    = 9/10 # = P(A). 9 invoices have item Bread out of 10\n",
    "C_A_B   =  S_A_B / P1 \n",
    "print('Confidence(Bread a∩d Milk)  is %1.4f' % C_A_B) "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Lift is a measure for a rule.\n",
    "\n",
    "* With lift value you can interpret the importance of a rule. \n",
    "* The lift value of an association rule is the ratio of the confidence of the rule and the expected confidence of the rule.\n",
    "* The expected confidence of a rule is the product of the support values of the rule body and rule head divided by the support of the rule body.\n",
    "* In our example, A is the rule head and B is the rule body.\n",
    "* Lift = $\\frac {P(A ∩ B)} {P(A) P(B)}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Assume L_A_B be the Lift of purchasing a item B\n",
      "\tgiven the item A is purchased.\n",
      "S_A_B =  P(A ∩ B) as discussed earlier.\n",
      "P1    = 0.9 as shown above\n",
      "Lift(Bread a∩d Milk)  is 0.8333\n"
     ]
    }
   ],
   "source": [
    "print('Assume L_A_B be the Lift of purchasing a item B') \n",
    "print('\\tgiven the item A is purchased.')\n",
    "print('S_A_B =  P(A ∩ B) as discussed earlier.')\n",
    "print('P1    = 0.9 as shown above')\n",
    "P2      =  4 / 10 # P(B) Refer invoices 1,5,6 and 10\n",
    "L_A_B   =  S_A_B / (P1  * P2)\n",
    "print('Lift(Bread a∩d Milk)  is %1.4f' % L_A_B) "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Association rules can be generated based on threshold values of support, confidence and lift.\n",
    "Assume that the threshold values are given for these measures as follows:\n",
    "* Support    = 0.25\n",
    "* Confidence = 0.50\n",
    "* Lift       > 1\n",
    "\n",
    "For the rule Bread >> Milk, \n",
    "* support    is 0.30\n",
    "* confidence is 0.33\n",
    "* lift       is 0.83\n",
    "\n",
    "So, this rule is not qualified to be considered for the association rule."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "A lift smaller than 1 indicates that the item A and item B appear less often together than expected.\n",
    "This implies that the occurrence of item A has a negative effect on the occurrence of item B."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Bayes theorem"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Bayes' Theorem is used to revise previously calcualted probabilities based on new information"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$P(B_{i}\\mid A)$ = $\\frac{P(A \\mid B_{i}) P(B_{i})}{P(A \\mid B_{1})P(B_{1}) + P(A \\mid B_{2}) P(B_{2}) + P(A \\mid B_{3}) P(B_{3}) + .. + P(A \\mid B_{k}) P(B_{k})}$\n",
    "\n",
    "where \n",
    "$B_{i}$ is the ith event of k mutually exclusive and collectively exhaustive events\n",
    "A is the new event that might impact P($B_{i}$)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 9"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "A certain Electronic equipment is manufactured by three companies, X, Y and Z. \n",
    "* 75% are manufactured by X\n",
    "* 15% are manufactured by Y\n",
    "* 10% are manufactured by Z\n",
    "\n",
    "The defect rates of electronic equipement manufactured by companies X, Y and Z are 4%, 6% and 8%.\n",
    "\n",
    "If an electronic equipment is randomly found to be defective, what is the probability that it is manufactured by X?"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "* Let P(X),P(Y) and P(Z) be probabilities of the electronic equipment manufactured by companies X, Y and Z respectively. \n",
    "* Let P(D) be the probability of defective electronic equipment.\n",
    "\n",
    "We are interested in calculating the probability P(X|D).\n",
    "\n",
    "P(X|D) = $\\frac{P(D | X) P(X)} {P(D)}$\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Bayes' rule in our case is given below:\n",
    "\n",
    "$P(X \\mid D)$ = $\\frac{P(D \\mid X) P(X)} {P(D \\mid X)P(D) + P(D \\mid Y) P(D) + P(D \\mid Z) P(D)}$\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "P(X|D)  is 0.6383\n"
     ]
    }
   ],
   "source": [
    "# Let P1 = P(D|X)\n",
    "# Let P2 = P(X)\n",
    "# Let P3 = P(D∣X)P(D)+P(D∣Y)P(D)+P(D∣Z)P(D)\n",
    "# Let P  = P(X|D) = (P1 * P2) / P3\n",
    "\n",
    "P1       =  0.04 # prob. of defective item manufactured by X\n",
    "P2       =  0.75\n",
    "P3       =  0.75 * 0.04 + 0.15 * 0.06 + 0.10 * 0.08 \n",
    "P        =  round((P1 * P2)/P3,4)\n",
    "print('P(X|D)  is %1.4f' % P)              "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 10"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Given the following statistics, what is the probability that a women has cancer if she has a positive mammogram result?\n",
    "\n",
    "a) \t1% of over 50 have breast cancer.\n",
    "b) \t90% of women who have breast cancer test positive on mammograms,\n",
    "c) \t8% of women will have false positive\n",
    "\n",
    "Let \n",
    "* event A   denote woman has breast cancer\n",
    "* event ~A  denote woman has no breast cancer\n",
    "* event T   denote mammogram test is positive \n",
    "* event ~T  denote mammogram test is negative \n",
    "\n",
    "Let P(A) denote the probability of women over 50 years of age having breast cancer.\n",
    "P(A)      = 0.01 \n",
    "So, P(~A) = 1 - 0.01 = 0.99\n",
    "\n",
    "Let P(T|A) denote the conditional probability of women given postive result on mammograms and  having breast cancer .\n",
    "\n",
    "P(T|A)   = 0.9\n",
    "\n",
    "Let P(T|~A) denote the conditional probability of women given the positive result on mammograms and not having breast cancer .\n",
    "\n",
    "P(T|~A)  = 0.08\n",
    "\n",
    "$P(T \\mid A)$ = $\\frac{P(A \\mid T) P(A)} {P(A \\mid T)P(A) + P(~A \\mid T) P(T)}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The probability of a women having cancer, given a positive test result is 0.1020\n"
     ]
    }
   ],
   "source": [
    "## Let P = P(T∣A)\n",
    "P = (0.9 * 0.01) / ((0.9 * 0.01) + (0.08 * 0.99)) \n",
    "print('The probability of a women having cancer, given a positive test result is %1.4f' % P) "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 11"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "A1 Electronic World is considering marketing a new model of televisions. In the past, 40% of the new model televisions have been successful, and 60% have been unsuccessful. Before introducing the new model television, the marketing research department conducts an extensive study and releases a report, either favorble or unfavorable.\n",
    "\n",
    "In the past, 80% of the successful new-model televisions had received favorable market reports, and 30% of the unsuccessful new-model televisions had received favorable reports. For the new model of television under consideration, the marketing research department has issued a favorable report.\n",
    "\n",
    "**What is the probability that the television is successful given a favorable report?**"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Let the following events be represented as follows:\n",
    "* S  denote the successful television\n",
    "* US denote the unsuccessful television\n",
    "* F  denote the favorable report\n",
    "* UF denote the unfavorable television\n",
    " \n",
    " The equation (Bayes' Theorem ) for this problem is:"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$P(S\\mid F)$ = $\\frac{P(F\\mid S)P(S)}{P(F\\mid S)P(S)) + P(F\\mid US)P(US))}$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Prior Probability** - - Revealed by data in the past\n",
    "* P(S)  = 0.40\n",
    "* P(US) = 0.60"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Conditional Probability**\n",
    "\n",
    "* $P(F\\mid S)$   = 0.80    ** Favorable Report received when the new model is successful**\n",
    "* $P(F\\mid US)$  = 0.20    ** Favorable Report received when the new model is unsuccessful**"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Using the equation given above,\n",
    "\n",
    "$P(S\\mid F)$ = $\\frac{P(F\\mid S)P(S)}{P(F\\mid S)P(S)) + P(F\\mid US)P(US))}$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$P(S\\mid F)$ = $\\frac{(0.80) (0.40)}{(0.80)(0.40) + (0.30)(0.60)}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The probability that the television is successful given a favorable report is 0.6400\n"
     ]
    }
   ],
   "source": [
    "# Let P = P(S∣F)\n",
    "P = (0.80 * 0.40)/ ((0.80 * 0.40) + (0.30 * 0.60))\n",
    "print('The probability that the television is successful given a favorable report is %1.4f' % P) "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### In class LAB :Practice Exercise 5"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "** The following contingency table list the probabilities of a person having a certain disease and the result of the diagnostic test conducted to determine whether the person is having the disease.**\n",
    "\n",
    "Let D be the event of having the disease and the event D' not having the disease.\n",
    "Let T be the event of having the test result positive and T' be the event of not having the test result positive.\n",
    "\n",
    "P(D)    = 0.03\n",
    "P(D')   = 0.97\n",
    "P(T|D)  = 0.90\n",
    "P(T|D') = 0.01\n",
    "\n",
    "What is the probability that the disease is acually present given a positive test result?\n",
    "\n",
    "**Hint: Use Bayes theorem**"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Take Home Exercise"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**1) A1 Construction company is determining whether it should submit a bid for  a new shopping mall. In the past, A1's main competitor, Pyramid Construction company has submitted bids 70% of the time. If  Pyramid Construction company does not bid on a job, the probability that A1 Construction Company will get the job is 0.50.** \n",
    "\n",
    "If the Pyramid Construction company bids on a job, the probability that A1 Construction company will get the job is 0.25.\n",
    "\n",
    "* a. If A1 Construction company gets the job, what is the probability that Pyramid Construction company did not bid?\n",
    "\n",
    "* b. What is the probability that A1 Construction company will get the job?"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Hint: Use Bayes theorem"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**2) From the table in Exercise 4, answer the following:**\n",
    "* a. What is the probability of Loan non-defaulted?\n",
    "* b. What is the conditional probability of non-default, given the occupation category- Employed in Government Sector?"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**3) You flip a coin and roll a six-sided dice, what is the probability that the coin comes up tail and the dice comes up with 6?**"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### End of Chapter 1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Chapter 2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Binomial and normal distribution"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "* It is widely used probability distribution of a discrete random variable. \n",
    "* Plays major role in quality control and quality assurance function. "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$P(X = x\\mid n,\\pi)$ = $\\frac{n!}{x!(n - x)!}\\pi^x (1 - \\pi)^{n-x} $\n",
    "* where P(X = x) is the probability of getting x successes in n trials\n",
    "and $\\pi$ is the probability of an event of interest"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Some important functions in Python for Binomial distribution:\n",
    "\n",
    "#### 1) Probability mass function\n",
    "\n",
    "##### scipy.stats.binom.pmf gives the probability mass function for the binomial distribution\n",
    "\n",
    "binomial = scipy.stats.binom.pmf (k,n,p), \n",
    "* where k is an array and takes values in {0, 1,..., n}\n",
    "* n and p are shape parameters for the binomial distribution\n",
    "\n",
    "The output, binomial, gives probability of binomial distribution function in terms of array.\n",
    "\n",
    "#### 2) Cumulative Density function\n",
    "\n",
    "cumbinomial = scipy.stats.binom.cdf(k,n,p) gives cumulative binomial distribution.\n",
    "\n",
    "The output, cumbinomial, gives cumulative probability of binomial distribution function in terms of array."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "####  3)  Plot the binomial Density function\n",
    "\n",
    "The function, matplotlib.pyplot.plot(k, binomial, ‘o-’) gives us plot of the binomial distribution function."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "A LED bulb manufacturing company regularly conducts quality checks at specified periods on the products it manufactures.\n",
    "Historically, the failure rate for LED light bulbs that the company manufactures is 5%. Suppose a random sample of 10 LED light bulbs is selected.\n",
    "What is the probability that \n",
    "* a) None of the LED bulbs are defective?\n",
    "* b) Exactly one of the LED bulbs is defective?\n",
    "* c) Two or fewer of the LED bulbs are defective?\n",
    "* d) Three or more of the LED bulbs are defective"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "import numpy             as np\n",
    "import pandas            as pd\n",
    "import scipy.stats       as stats\n",
    "import matplotlib.pyplot as plt"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "p   =  0.05 # failure rate for LED light bulbs that the company manufactures is 5%\n",
    "n   =  10 # sample size\n",
    "k   =  np.arange(0,11) # An array of quantiles"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "binomial = stats.binom.pmf(k,n,p)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[  5.98736939e-01   3.15124705e-01   7.46347985e-02   1.04750594e-02\n",
      "   9.64808106e-04   6.09352488e-05   2.67259863e-06   8.03789063e-08\n",
      "   1.58642578e-09   1.85546875e-11   9.76562500e-14]\n"
     ]
    }
   ],
   "source": [
    "print(binomial)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "a) None of the LED bulbs are defective? is 0.5987\n"
     ]
    }
   ],
   "source": [
    "print('a) None of the LED bulbs are defective? is %1.4f' %binomial[0])"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "b) Exactly one of the LED bulbs is defective? is 0.3151\n"
     ]
    }
   ],
   "source": [
    "print('b) Exactly one of the LED bulbs is defective? is %1.4f' %binomial[1])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "To answer the question c) Two or fewer of the LED bulbs are defective?\n",
    "we need to calculate cumulative probability of Upto Two LED bulbs being defective"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "cumbinomial = stats.binom.cdf(k,n,p)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[ 0.59873694  0.91386164  0.98849644  0.9989715   0.99993631  0.99999725\n",
      "  0.99999992  1.          1.          1.          1.        ]\n"
     ]
    }
   ],
   "source": [
    "print(cumbinomial)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "c) Two or fewer of the LED bulbs are defective? is 0.9885\n"
     ]
    }
   ],
   "source": [
    "print('c) Two or fewer of the LED bulbs are defective? is %1.4f' %cumbinomial[2])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### To answer the question d) Three or more of the LED bulbs are defective, we need to subtract cumulative Probability upto 2 defective LED bulbs from 1."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "d) Three or more of the LED bulbs are defective is 0.0115\n"
     ]
    }
   ],
   "source": [
    "P = 1- cumbinomial[2]\n",
    "print('d) Three or more of the LED bulbs are defective is %1.4f' % P)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Plot the binomial distribution function"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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hpBpJNevX+7HL1uzTrw8fOHwUdz71Ktt3u0iembVfrqRwaEScHxE/JTO4znva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\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x20bddb02d30>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "plt.plot(k,binomial, 'o-')\n",
    "plt.title('Binomial')\n",
    "plt.xlabel('Number of Def LED bulbs')\n",
    "plt.ylabel('Prob of Defective LED bulbs')\n",
    "plt.show()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The percentage of orders filled correctly at Wendy's was approximately 86.8%. Suppose that you go to drive-through window at Wendy's and place an order. Two friends of yours independently place orders at the drive-through window at the same Wendy's.\n",
    "\n",
    "What are the probabilities that \n",
    "* a) all three \n",
    "* b) none of the three\n",
    "* c) atleast two of the three orders will be filled correctly?\n",
    "* d) what is the mean and standard deviation of the orders filled correctly?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "p   =  0.868 # percentage of orders filled correctly at Wendy's was approximately 86.8%\n",
    "n   =  3\n",
    "k   =  np.arange(0,4)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "binomial = stats.binom.pmf(k,n,p)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[ 0.00229997  0.0453721   0.2983559   0.65397203]\n"
     ]
    }
   ],
   "source": [
    "print(binomial)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "a) all the three orders are filled correctly? is 0.6540\n"
     ]
    }
   ],
   "source": [
    "print('a) all the three orders are filled correctly? is %1.4f' %binomial[3])"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "b) none of the three orders are filled correctly? is 0.0023\n"
     ]
    }
   ],
   "source": [
    "print('b) none of the three orders are filled correctly? is %1.4f' %binomial[0])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "To answer this question c) atleast two of the three orders are filled correctly, \n",
    "we need to find out 1 - Probability of upto 1 failure"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "cumbinomial = stats.binom.cdf(k,n,p)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[ 0.00229997  0.04767206  0.34602797  1.        ]\n"
     ]
    }
   ],
   "source": [
    "print(cumbinomial)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "c) atleast two of the three orders are filled correctly? is 0.9523\n"
     ]
    }
   ],
   "source": [
    "P = 1 - cumbinomial[1]\n",
    "print('c) atleast two of the three orders are filled correctly? is %1.4f' % P)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "mean and standard deviation of the orders filled correctly are 2.6040 and  0.5863 respectively\n"
     ]
    }
   ],
   "source": [
    "# d) what is the mean and standard deviation of the orders filled correctly?\n",
    "p    = 0.868\n",
    "n    = 3\n",
    "mean = n* p # = 3 * 0.868\n",
    "sd   = np.sqrt(n * p * (1 - p))\n",
    "print('mean and standard deviation of the orders filled correctly are %1.4f and  %1.4f respectively' % (mean, sd))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Plot  the binomial distribution function"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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dzXo4/Pezcm67pOfwvcSCZiOADODekpa7+1PunuHuGW3atClnOCKJk5cfVBdd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\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x20bde01b898>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "plt.plot(k,binomial, 'o-')\n",
    "plt.title('Binomial')\n",
    "plt.xlabel('Number of Orders filled correctly')\n",
    "plt.ylabel('Prob of Orders filled correctly')\n",
    "plt.show()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### In class LAB : Practice Exercise 1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "** You and your two friends go to Pizza Hut which recently filled approximately 88.3% of the orders correctly.\n",
    "What is the probability that**"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "* a. all three orders will be filled correctly?\n",
    "* b. none of orders will be filled correctly?\n",
    "* c. at least two of three orders will be filled correctly?\n",
    "* d. What are the mean and standard deviation of the orders filled correctly?\n",
    "\n",
    "**Hint:** \n",
    "Mean and standard deviation of binomial distribution:\n",
    "* $\\mu     = n\\pi$ \n",
    "* $\\sigma  = \\sqrt{n\\pi(1-\\pi)}$ \n",
    "where n is the size of the sample and $\\pi$ is the probability of the event of interest"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Poisson Distribution"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "* This discrete distribution which also plays a major role in quality control. \n",
    "* The Poisson distribution is a discrete probability distribution for the counts of events that occur randomly in a given interval of time or space. In such areas of opportunity, there can be more than one occurrence. In such situations, Poisson distribution can be used to compute probabilities.\n",
    "* Examples include number of defects per item, number of defects per transformer produced. \n",
    "* Notes: Poisson Distribution helps to predict the arrival rate in a waiting line situation where a queue is formed and people wait to be served and the service rate is generally higher than the arrival rate.\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Properties:\n",
    "* Mean                     μ  = λ\n",
    "* Standard deviation       σ  = √ μ\n",
    "* The Poisson distribution is the limit of binomial distribution as n approaches ∞and p approaches 0\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "P(X = x) = $\\frac{e^\\lambda \\lambda^x}{x!} $\n",
    "where \n",
    "* P(x)              = Probability of x successes given an idea of  $\\lambda$\n",
    "* $\\lambda$ = Average number of successes\n",
    "* e                   = 2.71828 (based on natural logarithm)\n",
    "* x                    = successes per unit which can take values 0,1,2,3,... $\\infty$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Applications\n",
    "\n",
    "1. Car Accidents\n",
    "2. Number of deaths by horse kicking in Prussian Army (first application)\n",
    "3. Birth defects and genetic mutation\n",
    "\n",
    "### Note\n",
    "\n",
    "* If there is a fixed number of observations, n, each of which is classified as an event of interest or not an event of interest, use the binomial distribution.\n",
    "* If there is an area of opportunity, use the Poisson distribution.\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Some important functions in Python for solving Poisson distribution problems\n",
    "\n",
    "#### 1) Probability Mass Function\n",
    "\n",
    "poisson =  scipy.stats.poisson.pmf(n, rate) where n is where n is an array like quantiles and rate is the mean \n",
    "It gives poisson distribution result in the form of an array.\n",
    "\n",
    "#### 2) Cumulative Density Function\n",
    "\n",
    "poisson =  scipy.stats.poisson.cdf(n,rate) where n is where n is an array like quantiles and rate is the mean \n",
    "It gives cumulative density function result in the form of an array.\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 3"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The number of work-related injuries per month in a manufacturing plant is known to follow a Poisson distribution, with a mean  of 2.5 work-related injuries a month. What is the probability that in a given month, \n",
    "a) No work related injuries occur? \n",
    "b) At least one work-related injury occurs?"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Here, \n",
    "$\\lambda$ = 2.5 injuries"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 22,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "rate =  2.5 # which is the mean\n",
    "n    =  np.arange(0,20)\n",
    "poisson = stats.poisson.pmf(n,rate)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([  8.20849986e-02,   2.05212497e-01,   2.56515621e-01,\n",
       "         2.13763017e-01,   1.33601886e-01,   6.68009429e-02,\n",
       "         2.78337262e-02,   9.94061650e-03,   3.10644266e-03,\n",
       "         8.62900738e-04,   2.15725184e-04,   4.90284510e-05,\n",
       "         1.02142606e-05,   1.96428089e-06,   3.50764445e-07,\n",
       "         5.84607408e-08,   9.13449075e-09,   1.34330746e-09,\n",
       "         1.86570481e-10,   2.45487475e-11])"
      ]
     },
     "execution_count": 23,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "poisson"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " a) Probability that No work related injuries occur is 0.0821\n"
     ]
    }
   ],
   "source": [
    "print(' a) Probability that No work related injuries occur is %1.4f' % poisson[0])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "To calculate the probability that at least one work-related injury occurs, we need to subtract probability of no work related injury from 1."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 25,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " b) Probability that that at least one work-related injury occurs is 0.9179\n"
     ]
    }
   ],
   "source": [
    "# P = probability that at least one work-related injury occurs\n",
    "P   = 1 - poisson[0]\n",
    "print(' b) Probability that that at least one work-related injury occurs is %1.4f' % P)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Plot the Poisson distribution function¶"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 26,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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OQuC2cF3aVFbX0L0gjxH9vSdUIklMn1jG/DXb2XugPtvhOOeyoNlRZ83sCeAJ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\nQmCC\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x20bdf13bb70>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "plt.plot(n,poisson, 'o-')\n",
    "plt.title('Poisson')\n",
    "plt.xlabel('Number of Work related injuries')\n",
    "plt.ylabel('Prob of Work related injuries')\n",
    "plt.show()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 4"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "A Life Insurance agent sells on the average 3 life insurance policies per week. \n",
    "Use the Poisson law to calculate the probability that in a given week, he will sell\n",
    "a. Some policies\n",
    "b. 2 or more but less than 5 policies?"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Here $\\lambda$ = 3"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 27,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "rate =  3 # which is the mean\n",
    "n    =  np.arange(0,16)\n",
    "cumpoisson = stats.poisson.cdf(n,rate)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 28,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([ 0.04978707,  0.19914827,  0.42319008,  0.64723189,  0.81526324,\n",
       "        0.91608206,  0.96649146,  0.9880955 ,  0.99619701,  0.99889751,\n",
       "        0.99970766,  0.99992861,  0.99998385,  0.9999966 ,  0.99999933,\n",
       "        0.99999988])"
      ]
     },
     "execution_count": 28,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "cumpoisson"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 29,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " a) Probability that the agent sells some policies is 0.9502\n"
     ]
    }
   ],
   "source": [
    "# P = probability that the agent sells some policies\n",
    "P   = 1 - cumpoisson[0]\n",
    "print(' a) Probability that the agent sells some policies is %1.4f' % P)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    " b. Agent sells 2 or more but less than 5 policies "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 30,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " b) Probability that the agent sells between 2 and 5 policies is 0.6161\n"
     ]
    }
   ],
   "source": [
    "# Let P1 be the probability that the agent sells more than 4 policies\n",
    "P1 = cumpoisson[4] # P(X >=4)\n",
    "# Let P2 be the probability that the agent sells more than 1 policy\n",
    "P2 = cumpoisson[1] # P(X >=1)\n",
    "P = P1 - P2 # Prob. that the agent will sell between 2 and 5 policies\n",
    "print(' b) Probability that the agent sells between 2 and 5 policies is %1.4f' % P)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Plot the Poisson distribution function"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 31,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "data": {
      "image/png": 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ShoaoA+qcc20pSmHxaYL05I+Etz5A3uSNakweONn7K2Ijic+fMYIVm3YyZ7mv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\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x20bdf1ae470>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "poisson = stats.poisson.pmf(n,rate)\n",
    "plt.plot(n,poisson, 'o-')\n",
    "plt.title('Poisson')\n",
    "plt.xlabel('Number of Policies sold per week')\n",
    "plt.ylabel('Cumulative Prob of Policies sold per week')\n",
    "plt.show()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### In class LAB :Practice Exercise 2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "** Refer to Example 4. What is the probability that the Agent will sell**\n",
    "* a. At least 4 policies\n",
    "* b. Exactly 4 policies"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Normal Distribution"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "* One of the most popular continuous distribution in Analytics field.\n",
    "* Normal distribution is observed across many naturally occurring measures such as birth weight, height and intelligence etc"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Probability Density Function"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "f(x) = $\\frac {1}{\\sigma\\sqrt(2 \\pi)} e^\\frac {-(x-\\mu^2)}{2\\sigma^2}$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Where\n",
    "* f(x) is used to represent a probability density function\n",
    "* x is any value of the continuous variable, where -∞ < x < ∞\n",
    "* e denotes the mathematical constant approximated by 2.71828\n",
    "* Π is a mathematical constant approximated by 3.14159\n",
    "* μ and σ are the mean and standard deviation of the normal distribution\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "For a continuous function, the probability density function is the probability that the value has the value x. \n",
    "Since for continuous distributions the probability at a single point is zero, this is expressed in terms of an integral of its probability density function\n",
    "P(X<= x) = F(x) = \n",
    "$\\int_{-\\infty}^xf(t)dt$\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Standardize normal variable\n",
    "Compute Z by subtracting the mean, mu from a normally distributed variable, divide by the standard deviation, sigma.\n",
    "Z = (X - μ) / σ \n",
    "Z is in standard units. \n",
    "Z ~ N(0,1) and the variable, Z always has mean 0 and standard deviation 1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Its pdf is given by\n",
    "\n",
    "fZ(z) = $\\frac{1}{ √2Π} {exp \\frac {-z2} {2}}$\n",
    "\n",
    "for all z $\\in R$\n",
    "\n",
    "The (1 / √2Π ) is there to make sure that the area under the PDF is 1."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "* For a normal distribution, mu is the location parameter, which locates (centre) the distribution on the horizontal axis.\n",
    "* Sigma is the scale parameter, which defines the spread of the normal distribution.\n",
    "* Normal distribution has no shape parameter since all the normal distribution curves have bell shape and are symmetrical. \n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Properties\n",
    "1. Theoretical normal density functions are defined between -∞ and ∞\n",
    "2. There are two parameters, location (μ which is the mean) and scale (σ which is standard deviation).\n",
    "3. It has a symmetrical (bell shape) around the mean. mean = median = mode\n",
    "4. Areas between specific values are measured in terms of μ and σ\n",
    "5. Any linear transformation if a normal random variable is also normal random variable.\n",
    "6. If X1 is an independent normal random variable with mean μ1 and variance $\\sigma1^2$ and \n",
    "      X2 is another independent normal random variable with mean μ2 and$\\sigma2^2$, \n",
    "then X1 + X2 is also a normal distribution with mean μ1 + μ2 and variance $\\sigma1^2$ + $\\sigma2^2$ "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "| Value of the random variable | Area under the normal distribution (CDF) |\n",
    "| --------------------------------- | --------------------------|\n",
    "| $\\mu - \\sigma \\leq x \\leq \\mu + \\sigma$ | 0.6828 |\n",
    "| $\\mu - 2\\sigma \\leq x \\leq \\mu + 2\\sigma$ | 0.9545 |\n",
    "| $\\mu - 3\\sigma \\leq x \\leq \\mu + 3\\sigma$ | 0.9974 |"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Some important functions in python for solving Normal Distribution problems\n",
    "\n",
    "#### 1. Cumulative Density Function (cdf)\n",
    "\n",
    "1. scipy.stats.norm.cdf(z)                 # Here z is an attribute\n",
    "\n",
    "2. stats.norm.cdf(z2) – stats.norm.cdf(z1) # Here z is an attribute\n",
    "\n",
    "3. stats.norm.isf(0.99) # Inverse Survival function gives the value given a probability"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 5"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "A survey on use of smart phones in India was conducted and it is observed the smart phone users spend 68 minutes in a day on average in sending messages and the corresponding standard deviation is 12 minutes. \n",
    "\n",
    "* Assume that the time spent in sending messages follows a normal distribution. \n",
    "* a) What proportion of the smart phone users are spending more than 90 minutes in sending messages daily?\n",
    "* b) What proportion of customers are spending less than 20 minutes?\n",
    "* c) What proportion of customers are spending between 50 minutes and 100 minutes?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 32,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "mu    = 68 # minutes \n",
    "sigma = 12"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Compute Z by subtracting the mean, mu from a normally distributed variable, divide by the standard deviation, sigma"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 33,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " a. Proportion of the smart phone users are spending more than 90 minutes in sending messages daily is 0.0334\n"
     ]
    }
   ],
   "source": [
    "# a. What proportion of the smart phone users are spending more than 90 minutes in sending messages daily?\n",
    "z  = (90 - mu) / sigma # X = 90\n",
    "P  = 1 - stats.norm.cdf(z)\n",
    "print(' a. Proportion of the smart phone users are spending more than 90 minutes in sending messages daily is %1.4f' % P)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 34,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " b) What proportion of customers are spending less than 20 minutes is 0.0000317\n"
     ]
    }
   ],
   "source": [
    "# b) What proportion of customers are spending less than 20 minutes?\n",
    "z1  =  (20 - mu) / sigma # X = 20\n",
    "P   =  stats.norm.cdf(z1)\n",
    "print(' b) What proportion of customers are spending less than 20 minutes is %1.7f' % P)\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 35,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " c) proportion of customers are spending between 50 minutes and 100 minutes is 0.9294\n"
     ]
    }
   ],
   "source": [
    "# c) What proportion of customers are spending between 50 minutes and 100 minutes?\n",
    "z1  =  (50 - mu) / sigma   # X = 50\n",
    "z2  =  (100 - mu) / sigma  # X = 100\n",
    "P1   =  stats.norm.cdf(z1)\n",
    "P2   =  stats.norm.cdf(z2)\n",
    "P    =  P2 - P1\n",
    "print(' c) proportion of customers are spending between 50 minutes and 100 minutes is %1.4f' % P)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 6"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The mean salaries of Data Scientists working in Chennai, India is calculated to be 7,00,000 INR with a standard deviation of 90,000 INR. The random variable salary of Data Scientists follows a normal distribution.\n",
    "\n",
    "* a) What is the probability that a Data Scientist in Chennai has a salary more than 10,00,000 INR?\n",
    "* b) What is the probability that a Data Scientist in Chennai has a salary between 6,00,000 & 9,00,000 INR?\n",
    "* c) What is the probability that a Data Scientist in Chennai has a salary less than 4,00,000 INR?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 36,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "# a) What is the probability that a Data Scientist in Chennai has a salary more than 10,00,000 INR?\n",
    "mu    =  700000\n",
    "sigma =  90000\n",
    "z=(1000000-mu) / sigma "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 37,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " a) Probability that a Data Scientist in Chennai has a salary more than 10,00,000 is 0.0004\n"
     ]
    }
   ],
   "source": [
    "P = 1 - stats.norm.cdf(z)\n",
    "print(' a) Probability that a Data Scientist in Chennai has a salary more than 10,00,000 is %1.4f' % P)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 38,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "# b) What is the probability that a Data Scientist in Chennai has a salary between 6,00,000 & 9,00,000 INR?\n",
    "z1= (600000-mu) / sigma\n",
    "z2= (900000-mu) / sigma"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 39,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " b) probability that a Data Scientist in Chennai has a salary between 6,00,000 & 9,00,000 INR is 0.8536\n"
     ]
    }
   ],
   "source": [
    "P = stats.norm.cdf(z2) - stats.norm.cdf(z1)\n",
    "print(' b) probability that a Data Scientist in Chennai has a salary between 6,00,000 & 9,00,000 INR is %1.4f' % P)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 40,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "# c) What is the probability that a Data Scientist in Chennai has a salary less than 4,00,000 INR?\n",
    "z = (400000-mu) / sigma "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 41,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " c) Probability that a Data Scientist in Chennai has a salary less than 4,00,000 INR is 0.0004\n"
     ]
    }
   ],
   "source": [
    "P = stats.norm.cdf(z)\n",
    "print(' c) Probability that a Data Scientist in Chennai has a salary less than 4,00,000 INR is %1.4f' % P)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 7"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The fill amount in 2-liter soft drink bottles is normally distributed, with a mean of 2.0 liters and a standard deviation of 0.05 liter. If the bottles contain less than 95% of the listed net content (1.90 liters, in our case), the manufacturer may be subject to penalty by the state office of consumer affairs. Bottles that have a net content above 2.1 liters may cause excess spillage upon opening. What is the proportion of bottles that will contain\n",
    "\n",
    "* a) between 1.9 and 2.0 liters\n",
    "* b) between 1.9 and 2.1 liters\n",
    "* c) below 1.9 liters or above 2.1 liters\n",
    "* d) At least how much soft drink is contained in 99% of the bottles?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 42,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "mu     = 2\n",
    "sigma  = 0.05"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### a. between 1.9 and 2.0 liters"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 43,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Z value for 1.90 liters is -2.0000\n"
     ]
    }
   ],
   "source": [
    "x1   = 1.90\n",
    "z1   = (x1 - mu) /  sigma \n",
    "print(\"Z value for 1.90 liters is %3.4f\" % z1)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 44,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Z value for 2 liters is 0.0000\n"
     ]
    }
   ],
   "source": [
    "x2   = 2\n",
    "z2   = (x2 - mu) /  sigma \n",
    "print(\"Z value for 2 liters is %3.4f\" % z2)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 45,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "a.  Proportion of bottles between 1.9 and 2 L is 0.4772\n"
     ]
    }
   ],
   "source": [
    "P  = stats.norm.cdf(z2) - stats.norm.cdf(z1)\n",
    "print(\"a.  Proportion of bottles between 1.9 and 2 L is %3.4f\" % P)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### b) between 1.9 and 2.1 liters"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 46,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Z value for 2.1 liters is 2.0000\n"
     ]
    }
   ],
   "source": [
    "x3   = 2.1\n",
    "z3   = (x3 - mu) /  sigma\n",
    "print(\"Z value for 2.1 liters is %3.4f\" % z3)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 47,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "b. Proportion of bottles between 1.9 and 2.1 L is 0.9545\n"
     ]
    }
   ],
   "source": [
    "P = stats.norm.cdf(z3) - stats.norm.cdf(z1)\n",
    "print(\"b. Proportion of bottles between 1.9 and 2.1 L is %3.4f\" %P)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### c. below 1.9 liters or above 2.1 liters"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 48,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "c. below 1.9 liters or above 2.1 liters is  0.0455\n"
     ]
    }
   ],
   "source": [
    "P1  =  stats.norm.cdf(z1)\n",
    "P2  =  1 - stats.norm.cdf(z3)\n",
    "print(\"c. below 1.9 liters or above 2.1 liters is  %3.4f\" % (P1 + P2))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### d) At least how much soft drink is contained in 99% of the bottles?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 49,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "d. At least how much soft drink is contained in 99 percent of the bottles is  1.8837 Liters\n"
     ]
    }
   ],
   "source": [
    "# Here we need to find the vol\n",
    "P = stats.norm.isf(q = 0.99, loc = mu, scale = sigma)\n",
    "print(\"d. At least how much soft drink is contained in 99 percent of the bottles is  %4.4f Liters\" % P)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "### In class LAB : Practice Exercise 3"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "** The following table contains the overall miles per Gallon (MPG) of 2017 small SUVs ( n =40).**\n",
    "* Decide whether the data appears to be normally distributed by\n",
    "* a. Mean and median "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[19 20 20 20 21 19 21 19 21 20 20 20 19 21 21 22 20 22 20 20 22 20 22 21 19\n",
      " 19 20 20 21 20 19 18 20 21 21 20 21 21 20 21]\n"
     ]
    }
   ],
   "source": [
    "mpg = np.array([19, 20, 20, 20, 21, 19, 21, 19, 21, 20, 20, 20, 19, 21, 21, 22, 20, 22, 20, 20, 22, 20, 22, 21, 19,\n",
    "                             19, 20, 20, 21, 20, 19, 18, 20, 21, 21, 20, 21, 21, 20,  21])\n",
    "print(mpg)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Hint: Use np.median(mpg) to calculate median and mpg.mean() to calculate mean"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 51,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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      "text/plain": [
       "<matplotlib.figure.Figure at 0x20bdf21b898>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "fig, ax = plt.subplots()\n",
    "pd.DataFrame(mpg).plot.kde(ax=ax, legend=False)\n",
    "pd.DataFrame(mpg).plot.hist(density=True, ax=ax)\n",
    "plt.show()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "* b. From the above plot, do you think the data is normally distributed?"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "|     |     |     |    | \n",
    "| --- | --- | --- | -- |\n",
    "| 19  | 20  | 20  | 20 |\n",
    "| 21  | 19  | 21  | 19 |\n",
    "| 21  | 20  | 20  | 20 |\n",
    "| 19  | 21  | 21  | 22 |\n",
    "| 20  | 22  | 20  | 20 |\n",
    "| 22  | 20  | 22  | 21 |\n",
    "| 19  | 19  | 20  | 20 |  \n",
    "| 21  | 20  | 19  | 18 | \n",
    "| 20  | 21  | 21  | 20 | \n",
    "| 21  | 21  | 20  | 21|\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### In class LAB :Practice Exercise 4"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**The mean weight of a morning breakfast cereal pack is 0.295 kg with a standard deviation of 0.025 kg. The random variable weight of the pack follows a normal distribution. **\n",
    "\n",
    "* a. What is the probability that the pack weighs less than 0.280 kg?\n",
    "* b. What is the probability that the pack weighs more than 0.350 kg? \n",
    "* c. What is the probability that the pack weighs between 0.260 kg to 0.340 kg? "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Take home exercises"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**1. If X is a random variable that follows normal distribution, then 5X + 40 is a**\n",
    "* a. Chi-square distribution\n",
    "* b. Poisson distribution\n",
    "* c. Binomial distribution\n",
    "* d. Normal distribution"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**2. The number of warranty claims received at an Electronic equipment manufacturer follows a Poisson distribution with a rate of 10 per day. ** \n",
    "Calculate the following:\n",
    "* a. The probability that the warranty claims is below 5 on a given day.\n",
    "* b. The probability that the warranty claims will exceed 10 on a given day."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**3. Refer to example 7. What is the proportion of bottles that will contain**\n",
    "* a) below 1.96 liters or above 2.2 liters\n",
    "* b) At least how much soft drink is contained in 90% of the bottles?"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### END"
   ]
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
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 },
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}