 Basic Statistics1- Ass2-Set4-Q3.ipynb

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  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Basic Statistics_Set4   ( Q3)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "3.\tAuditors at a small community bank randomly sample 100 withdrawal transactions made during the week at an ATM machine located near the bank’s main branch. Over the past 2 years, the average withdrawal amount has been $50 with a standard deviation of $40. Since audit investigations are typically expensive, the auditors decide to not initiate further investigations if the mean transaction amount of the sample is between $45 and $55. What is the probability that in any given week, there will be an investigation?\n",
    "\n",
    "A.\t1.25%\n",
    "\n",
    "B.\t2.5%\n",
    "\n",
    "C.\t10.55%\n",
    "\n",
    "D.\t21.1%\n",
    "\n",
    "E.\t50%\n"
   ]
  },
  {
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   "outputs": [],
   "source": [
    "import numpy as np\n",
    "from scipy import stats\n",
    "from scipy.stats import norm"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "n=100, Pop Mean=50, Pop SD=40, xbar1= 45, xbar2=55\n",
    "As no. of samples is more than 30, we can consider it normal distribution\n"
   ]
  },
  {
   "cell_type": "code",
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   "outputs": [],
   "source": [
    "# For No investigation P(45<X<55)\n",
    "# For Investigation 1-P(45<X<55)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
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   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "-1.25"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# find z-scores at x=45; z=(x1bar-P_mean)/(p_SD/sqrt(n))\n",
    "z1=(45-50)/(40/100**0.5)\n",
    "z1"
   ]
  },
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   "execution_count": 4,
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   "outputs": [
    {
     "data": {
      "text/plain": [
       "1.25"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# find z-scores at x=55; z=(xbar2-P_mean)/(p_SD/sqrt(n))\n",
    "z2=(55-50)/(40/100**0.5)\n",
    "z2"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
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   "outputs": [
    {
     "data": {
      "text/plain": [
       "0.7887004526662893"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# For No investigation P(45<X<55) using z_scores = P(X<55)-P(X<45)\n",
    "stats.norm.cdf(1.25)-stats.norm.cdf(-1.25)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
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   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "(45.00000495667348, 54.99999504332652)"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "stats.norm.interval(0.7887,loc=50,scale=40/(100**0.5)) #Loc=Population mean= 50"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
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   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "0.21130000000000004"
      ]
     },
     "execution_count": 7,
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   ],
   "source": [
    "# For Investigation 1-P(45<X<55)\n",
    "1-0.7887"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Answer -  Option D 21.1"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Basic Statistics_Set 4   ( Q4)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "4.\tThe auditors from the above example would like to maintain the probability of investigation to 5%. Which of the following represents the minimum number transactions that they should sample if they do not want to change the thresholds of 45 and 55? Assume that the sample statistics remain unchanged.\n",
    "\n",
    "A.\t144\n",
    "\n",
    "B.\t150\n",
    "\n",
    "C.\t196\n",
    "\n",
    "D.\t250\n",
    "\n",
    "E.\tNot enough information\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "n=100, Pop Mean=50, Pop SD=40, xbar1= 45, xbar2=55\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "###         1 - (45<P<55) = 0.05\n",
    "###          (45<P<55) = 0.95"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-08-16T06:40:15.444043Z",
     "start_time": "2021-08-16T06:40:15.277295Z"
    }
   },
   "outputs": [],
   "source": [
    "import numpy as np\n",
    "from scipy import stats\n",
    "from scipy.stats import norm"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-08-16T06:40:15.698339Z",
     "start_time": "2021-08-16T06:40:15.467981Z"
    }
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "(-1.959963984540054, 1.959963984540054)"
      ]
     },
     "execution_count": 9,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "stats.norm.interval(0.95)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-08-16T06:40:16.004760Z",
     "start_time": "2021-08-16T06:40:15.748207Z"
    }
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "243.35999999999999"
      ]
     },
     "execution_count": 10,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "n=((1.95*40)/(55-50))**2    #n = ((Z_55*sd)/(xbar2-mu))**2\n",
    "n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-08-11T16:28:19.924325Z",
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   },
   "source": [
    "n = 243.35 ≈ 250\n",
    "### Option D ,  250 is the right answer\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
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