{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 23 : Statistical Methods"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 23.1, page no. 618"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 65,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The first row of A denotes the no. of students falling in the marks group starting from (5 −10)... till (40−45)\n",
      "the second row denotes cumulative frequency (less than)\n",
      "the third row denotes cumulative frequency(more than)\n",
      "[[  5.   6.  15.  10.   5.   4.   2.   2.]\n",
      " [  5.  11.  26.  36.  41.  45.  47.  49.]\n",
      " [ 49.  44.  38.  23.  13.   8.   4.   2.]]\n"
     ]
    }
   ],
   "source": [
    "import numpy\n",
    "\n",
    "A = numpy.zeros([3,8])\n",
    "print \"The first row of A denotes the no. of students falling in the marks group starting from (5 −10)... till (40−45)\"\n",
    "A[0] = [5,6,15,10,5,4,2,2]\n",
    "print \"the second row denotes cumulative frequency (less than)\"\n",
    "A[1,0] = 5\n",
    "for i in range(1,8):\n",
    "    A[1,i] = A[1,i-1]+A[0,i]\n",
    "print \"the third row denotes cumulative frequency(more than)\"\n",
    "A[2,0] = 49\n",
    "for i in range (1,8):\n",
    "    A[2,i] = A[2,i-1]-A[0,i-1]\n",
    "print A"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 23.2, page no. 619"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 63,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The first row of A represents the mid values of weekly earnings having interval of 2 in each class=x \n",
      "The second row denotes the no. of employees or in other words frequency=f \n",
      "Third row denotes f∗x\n",
      "Fourth row denotes u=(x−25)/2 \n",
      "Fifth row denotes f∗x\n",
      "[[  1.10000000e+01   1.30000000e+01   1.50000000e+01   1.70000000e+01\n",
      "    1.90000000e+01   2.10000000e+01   2.30000000e+01   2.50000000e+01\n",
      "    2.70000000e+01   2.90000000e+01   3.10000000e+01   3.30000000e+01\n",
      "    3.50000000e+01   3.70000000e+01   3.90000000e+01   4.10000000e+01]\n",
      " [  3.00000000e+00   6.00000000e+00   1.00000000e+01   1.50000000e+01\n",
      "    2.40000000e+01   4.20000000e+01   7.50000000e+01   9.00000000e+01\n",
      "    7.90000000e+01   5.50000000e+01   3.60000000e+01   2.60000000e+01\n",
      "    1.90000000e+01   1.30000000e+01   9.00000000e+00   7.00000000e+00]\n",
      " [  3.30000000e+01   7.80000000e+01   1.50000000e+02   2.55000000e+02\n",
      "    4.56000000e+02   8.82000000e+02   1.72500000e+03   2.25000000e+03\n",
      "    2.13300000e+03   1.59500000e+03   1.11600000e+03   8.58000000e+02\n",
      "    6.65000000e+02   4.81000000e+02   3.51000000e+02   2.87000000e+02]\n",
      " [ -7.00000000e+00  -6.00000000e+00  -5.00000000e+00  -4.00000000e+00\n",
      "   -3.00000000e+00  -2.00000000e+00  -1.00000000e+00   0.00000000e+00\n",
      "    1.00000000e+00   2.00000000e+00   3.00000000e+00   4.00000000e+00\n",
      "    5.00000000e+00   6.00000000e+00   7.00000000e+00   8.00000000e+00]\n",
      " [ -2.10000000e+01  -3.60000000e+01  -5.00000000e+01  -6.00000000e+01\n",
      "   -7.20000000e+01  -8.40000000e+01  -7.50000000e+01   0.00000000e+00\n",
      "    7.90000000e+01   1.10000000e+02   1.08000000e+02   1.04000000e+02\n",
      "    9.50000000e+01   7.80000000e+01   6.30000000e+01   5.60000000e+01]]\n",
      "Sum of all elements of third row= 13315.0\n",
      "Sum of all elements of second row= 509.0\n",
      "mean= 26.1591355599\n",
      "Sum of all elements of fifth row=  295.0\n",
      "mean by step deviation method=  26.1591355599\n"
     ]
    }
   ],
   "source": [
    "import numpy\n",
    "\n",
    "A = numpy.zeros([5,16])\n",
    "print \"The first row of A represents the mid values of weekly earnings having interval of 2 in each class=x \"\n",
    "A[0] = [11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41]\n",
    "print \"The second row denotes the no. of employees or in other words frequency=f \"\n",
    "A[1]=[3,6,10,15,24,42,75,90,79,55,36,26,19,13,9,7]\n",
    "print \"Third row denotes f∗x\"\n",
    "for i in range(0,16):\n",
    "    A[2,i] = A[0,i]*A[1,i]\n",
    "print \"Fourth row denotes u=(x−25)/2 \"\n",
    "for i in range(0,16):\n",
    "    A[3,i] = (A[0,i]-25)/2\n",
    "print \"Fifth row denotes f∗x\"\n",
    "for i in range(0,16):\n",
    "    A[4,i] = A[3,i]*A[1,i]\n",
    "print A\n",
    "b = 0\n",
    "print \"Sum of all elements of third row=\",\n",
    "for i in range(0,16):\n",
    "    b = b+A[2,i]\n",
    "print b\n",
    "f = 0\n",
    "print \"Sum of all elements of second row=\",\n",
    "for i in range(0,16):\n",
    "    f = f+A[1,i]\n",
    "print f\n",
    "print \"mean=\",b/f\n",
    "d = 0\n",
    "print \"Sum of all elements of fifth row= \",\n",
    "for i in range(0,16):\n",
    "    d = d+A[4,i]\n",
    "print d\n",
    "print \"mean by step deviation method= \",25+(2*d/f)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 23.3, page no. 620"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 47,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The first row of A denotes the no. of students falling in the marks group startin from (5−10)... till (40−45)\n",
      "The second row denotes cumulative frequency (less than)\n",
      "The third row denotes cumulative frequency (more than)\n",
      "[[  5.   6.  15.  10.   5.   4.   2.   2.]\n",
      " [  5.  11.  26.  36.  41.  45.  47.  49.]\n",
      " [ 49.  44.  38.  23.  13.   8.   4.   2.]]\n",
      "Median falls in the class (15−20)=l+((n/2−c)∗h)/f=  19\n",
      "Lower quartile also falls in the class (15−20)=\n",
      "Upper quartile also falls in the class (25−30)=\n",
      "Semiinter quartile range=  5\n"
     ]
    }
   ],
   "source": [
    "import numpy\n",
    "\n",
    "A = numpy.zeros([3,8])\n",
    "print \"The first row of A denotes the no. of students falling in the marks group startin from (5−10)... till (40−45)\"\n",
    "\n",
    "A[0] = [5,6,15,10,5,4,2,2]\n",
    "print \"The second row denotes cumulative frequency (less than)\"\n",
    "A[1]=[5,11,26,36,41,45,47,49]\n",
    "print \"The third row denotes cumulative frequency (more than)\"\n",
    "A[2] = [49,44,38,23,13,8,4,2]\n",
    "print A\n",
    "print \"Median falls in the class (15−20)=l+((n/2−c)∗h)/f= \",15+((49/2-11)*5)/15\n",
    "print \"Lower quartile also falls in the class (15−20)=\"\n",
    "Q1 = 15+((49/4-11)*5)/15\n",
    "print \"Upper quartile also falls in the class (25−30)=\"\n",
    "Q3 =25+((3*49/4-36)*5)/5\n",
    "print \"Semiinter quartile range= \",(Q3-Q1)/2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 23.4, page no. 620"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 66,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The first row of A denotes the roll no. of students form 1 to 10 and that of B denotes form 11 to 20 \n",
      "The second row of A and B denotes the corresponding marks in physics\n",
      "The third row denotes the corresponding marks in chemistry\n",
      "Median marks in physics=arithmetic mean of 10th and 11th student =  26\n",
      "Median marks in chemistry=arithmetic mean of 10th and 11th student =  41\n"
     ]
    }
   ],
   "source": [
    "import numpy\n",
    "\n",
    "A = numpy.zeros([3,10])\n",
    "print \"The first row of A denotes the roll no. of students form 1 to 10 and that of B denotes form 11 to 20 \"\n",
    "A[0] = [1,2,3,4,5,6,7,8,9,10]\n",
    "B[0] = [11,12,13,14,15,16,17,18,19,20]\n",
    "print \"The second row of A and B denotes the corresponding marks in physics\"\n",
    "A[1] = [53,54,52,32,30,60,47,46,35,28]\n",
    "B[1] = [25,42,33,48,72,51,45,33,65,29]\n",
    "print \"The third row denotes the corresponding marks in chemistry\"\n",
    "A[2] = [58,55,25,32,26,85,44,80,33,72]\n",
    "B[2] = [10,42,15,46,50,64,39,38,30,36]\n",
    "print \"Median marks in physics=arithmetic mean of 10th and 11th student = \",(28+25)/2\n",
    "print \"Median marks in chemistry=arithmetic mean of 10th and 11th student = \",(72+10)/2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 23.5, page no. 621"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Let the misssing frequencies be f1 and f2 \n",
      "Sum of given frequencies=12+30+65+25+18 =  150\n",
      "So, f1+f2=229−c =  79\n",
      "Median=46=40+(114.5−(12+30+f1))∗10/65)\n",
      "f1=33.5=34 \n"
     ]
    }
   ],
   "source": [
    "print \"Let the misssing frequencies be f1 and f2 \"\n",
    "print \"Sum of given frequencies=12+30+65+25+18 = \",\n",
    "c =12+30+65+25+18\n",
    "print c\n",
    "print \"So, f1+f2=229−c = \",229-c\n",
    "print \"Median=46=40+(114.5−(12+30+f1))∗10/65)\"\n",
    "print \"f1=33.5=34 \"\n",
    "f1 = 34\n",
    "f2 = 45"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 23.6, page no. 622"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Let the eqidistance be s, then\n",
      "Average speed = total distance/total time taken 1800/47\n"
     ]
    }
   ],
   "source": [
    "import sympy\n",
    "\n",
    "s = sympy.Symbol('s')\n",
    "print \"Let the eqidistance be s, then\"\n",
    "t1 = s/30\n",
    "t2 = s/40\n",
    "t3 = s/50\n",
    "print \"Average speed = total distance/total time taken\",3*s/(t1+t2+t3)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 23.7, page no. 623"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 68,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The first row denotes the size of item \n",
      "The second row denotes the corresponding frequency(f)\n",
      "The third row denotes the corresponding deviation(d)\n",
      "The fourth row denotes the corresponding f∗d \n",
      "The fifth row denotes the corresponding f∗dˆ2 \n",
      "[[  6.   7.   8.   9.  10.  11.  12.]\n",
      " [  3.   6.   9.  13.   8.   5.   4.]\n",
      " [ -3.  -2.  -1.   0.   1.   2.   3.]\n",
      " [ -9. -12.  -9.   0.   8.  10.  12.]\n",
      " [ 27.  24.   9.   0.   8.  20.  36.]]\n",
      "Sum of fourth row elements=  0.0\n",
      "Sum of fifth row elements=  124.0\n",
      "Sum of all frequencies =  48.0\n",
      "mean=9+b/d =  9.0\n",
      "Standard deviation = (c/d)ˆ0.5 =  1.60727512683\n"
     ]
    }
   ],
   "source": [
    "import numpy\n",
    "\n",
    "A = numpy.zeros([5,7])\n",
    "print \"The first row denotes the size of item \"\n",
    "A[0,:] = [6,7,8,9,10,11,12]\n",
    "print \"The second row denotes the corresponding frequency(f)\"\n",
    "A[1,:] = [3,6,9,13,8,5,4]\n",
    "print \"The third row denotes the corresponding deviation(d)\"\n",
    "A[2,:] = [-3,-2,-1,0,1,2,3]\n",
    "print \"The fourth row denotes the corresponding f∗d \"\n",
    "for i in range(0,7):\n",
    "    A[3,i] = A[1,i]*A[2,i] \n",
    "print \"The fifth row denotes the corresponding f∗dˆ2 \"\n",
    "for i in range(0,7):\n",
    "    A[4,i] = A[1,i]*(A[2,i]**2)\n",
    "print A\n",
    "b = 0\n",
    "print \"Sum of fourth row elements= \",\n",
    "for i in range(0,7):\n",
    "    b = b+A[3,i]\n",
    "print b\n",
    "c = 0\n",
    "print \"Sum of fifth row elements= \",\n",
    "for i in range(0,7):\n",
    "    c = c+A[4,i]\n",
    "print c\n",
    "d = 0\n",
    "print \"Sum of all frequencies = \",\n",
    "for i in range(0,7):\n",
    "    d = d+A[1,i]\n",
    "print d\n",
    "print \"mean=9+b/d = \",9+b/d\n",
    "print \"Standard deviation = (c/d)ˆ0.5 = \",(c/d)**0.5"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 23.8, page no. 624"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 70,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The first row of A represents the mid values of wage classes having interval of 8 in each class=x \n",
      "The second row denotes the no. of men or in other words frequency=f \n",
      "Third row denotes f∗x \n",
      "Fourth row denotes d=(x−32.5)/8 \n",
      "Fifth row denotes f∗d \n",
      "Sixth row denotes f∗(dˆ2)\n",
      "[[   8.5   16.5   24.5   32.5   40.5   48.5   56.5   64.5   72.5]\n",
      " [   2.    24.    21.    18.     5.     3.     5.     8.     2. ]\n",
      " [  17.   396.   514.5  585.   202.5  145.5  282.5  516.   145. ]\n",
      " [  -3.    -2.    -1.     0.     1.     2.     3.     4.     5. ]\n",
      " [  -6.   -48.   -21.     0.     5.     6.    15.    32.    10. ]\n",
      " [  18.    96.    21.     0.     5.    12.    45.   128.    50. ]]\n",
      "Sum of all elements of sixth row=  375.0\n",
      "Sum of all elements of second row=  88.0\n",
      "mean=  4.26136363636\n",
      "Sum of all elements of fifth row=  Mean wage=  31.8636363636\n",
      "standard deviation=  34.0402892562\n"
     ]
    }
   ],
   "source": [
    "import numpy\n",
    "\n",
    "A = numpy.zeros([6,9])\n",
    "print \"The first row of A represents the mid values of wage classes having interval of 8 in each class=x \"\n",
    "A[0] = [8.5,16.5,24.5,32.5,40.5,48.5,56.5,64.5,72.5]\n",
    "print \"The second row denotes the no. of men or in other words frequency=f \"\n",
    "A[1] = [2,24,21,18,5,3,5,8,2]\n",
    "print \"Third row denotes f∗x \"\n",
    "for i in range(0,9):\n",
    "    A[2,i] = A[0,i]*A[1,i] \n",
    "print \"Fourth row denotes d=(x−32.5)/8 \"\n",
    "for i in range(0,9):\n",
    "    A[3,i] = (A[0,i]-32.5)/8\n",
    "print \"Fifth row denotes f∗d \"\n",
    "for i in range(0,9):\n",
    "    A[4,i] = A[3,i]*A[1,i]\n",
    "print \"Sixth row denotes f∗(dˆ2)\"\n",
    "for i in range(0,9):\n",
    "    A[5,i] = A[3,i]**2*A[1,i]\n",
    "print A\n",
    "b = 0\n",
    "print \"Sum of all elements of sixth row= \",\n",
    "for i in range(0,9):\n",
    "    b = b+A[5,i]\n",
    "print b\n",
    "f = 0\n",
    "print \"Sum of all elements of second row= \",\n",
    "for i in range(0,9):\n",
    "    f = f+A[1,i]\n",
    "print f\n",
    "print \"mean= \",b/f\n",
    "d = 0\n",
    "print \"Sum of all elements of fifth row= \",\n",
    "for i in range(0,9):\n",
    "    d = d+A[4,i]\n",
    "print \"Mean wage= \",32.5+(8*d/f)\n",
    "print \"standard deviation= \",8*(b/f-(d/f)**2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 23.9, page no. 626"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 72,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The first row of A denotes the scores of A and that of B denotes that of B \n",
      "The second row of A and B denotes the corresponding deviation \n",
      "The third row of A and B denotes the corresponding deviation square \n",
      "[[   12.   115.     6.    73.     7.    19.   119.    36.    84.    29.]\n",
      " [  -39.    64.   -45.    22.   -44.   -32.    68.   -15.    33.   -22.]\n",
      " [ 1521.  4096.  2025.   484.  1936.  1024.  4624.   225.  1089.   484.]]\n",
      "[[   47.    12.    16.    42.     4.    51.    37.    48.    13.     0.]\n",
      " [   -4.   -39.   -35.    -9.   -47.     0.   -14.    -3.   -38.   -51.]\n",
      " [   16.  1521.  1225.    81.  2209.     0.   196.     9.  1444.  2601.]]\n",
      "Sum of second row elements of  A=b= -10.0\n",
      "sum of second row elements of B=c= -240.0\n",
      "Sum of third row elements of A=d= 17508.0\n",
      "Sum of second row elements of B=e= 9302.0\n",
      "Arithmetic mean of A=  50.0\n",
      "Standard deviation of A=  41.8306108012\n",
      "Arithmetic mean of B=  27.0\n",
      "Standard deviation of A=  18.8202019118\n",
      "Coefficient of variation of A= 83.6612216024\n",
      "Coefficient of variation of B= 69.7044515251\n"
     ]
    }
   ],
   "source": [
    "import numpy\n",
    "\n",
    "A = numpy.zeros([3,10])\n",
    "B = numpy.zeros([3,10])\n",
    "print \"The first row of A denotes the scores of A and that of B denotes that of B \"\n",
    "A[0] = [12,115,6,73,7,19,119,36,84,29]\n",
    "B[0] = [47,12,16,42,4,51,37,48,13,0]\n",
    "print \"The second row of A and B denotes the corresponding deviation \"\n",
    "for i in range (0,10):\n",
    "    A[1,i] = A[0,i]-51\n",
    "    B[1,i] = B[0,i]-51\n",
    "print \"The third row of A and B denotes the corresponding deviation square \"\n",
    "for i in range (0,10):\n",
    "    A[2,i] = A[1,i]**2\n",
    "    B[2,i] = B[1,i]**2\n",
    "print A\n",
    "print B\n",
    "b = 0\n",
    "print \"Sum of second row elements of  A=b=\",\n",
    "for i in range (0,10):\n",
    "    b = b+A[1,i]\n",
    "print b\n",
    "c = 0\n",
    "print \"sum of second row elements of B=c=\",\n",
    "for i in range (0,10):\n",
    "    c = c+B[1,i]\n",
    "print c\n",
    "d = 0\n",
    "print \"Sum of third row elements of A=d=\",\n",
    "for i in range (0,10):\n",
    "    d = d+A[2,i]\n",
    "print d\n",
    "e = 0\n",
    "print \"Sum of second row elements of B=e=\",\n",
    "for i in range (0,10):\n",
    "    e = e+B[2,i]\n",
    "print e\n",
    "print \"Arithmetic mean of A= \",\n",
    "f = 51+b/10\n",
    "print f\n",
    "print \"Standard deviation of A= \",\n",
    "g = (d/10-(b/10)**2)**0.5\n",
    "print g\n",
    "print \"Arithmetic mean of B= \",\n",
    "h = 51+c/10\n",
    "print h\n",
    "print \"Standard deviation of A= \",\n",
    "i = (e/10-(c/10)**2)**0.5\n",
    "print i\n",
    "print \"Coefficient of variation of A=\",(g/f)*100\n",
    "print \"Coefficient of variation of B=\",(i/h)*100"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 23.10, page no. 628"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "If m is the mean of entire data, then\n",
      "116\n",
      "If s is the standard deviation of entire data, then \n",
      "7.74596669241\n"
     ]
    }
   ],
   "source": [
    "print \"If m is the mean of entire data, then\"\n",
    "m = (50*113+60*120+90*115)/(50+60+90)\n",
    "print m\n",
    "print \"If s is the standard deviation of entire data, then \"\n",
    "s = (((50*6**2)+(60*7**2)+(90*8**2)+(50*3**2)+(60*4**2)+(90*1**2))/200)**0.5\n",
    "print s"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 23.12, page no. 629"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 73,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The first row of A denotes the no. of persons falling in the weight group starting from (70−80)... till (140−150)\n",
      "The second row denotes cumulative frequency\n",
      "Median falls in the class (110−120) = l+((n/2−c)∗h)/f=  111\n",
      "Lower quartile also falls in the class (90−100)=  97.8571428571\n",
      "Upper quartile also falls in the class (120−130)=  123.444444444\n",
      "Quartile coefficient of skewness=  -0.0272952853598\n"
     ]
    }
   ],
   "source": [
    "import numpy\n",
    "\n",
    "A = numpy.zeros([2,8])\n",
    "print \"The first row of A denotes the no. of persons falling in the weight group starting from (70−80)... till (140−150)\"\n",
    "A[0] = [12,18,35,42,50,45,20,8]\n",
    "print \"The second row denotes cumulative frequency\"\n",
    "A[1,0] = 12\n",
    "for i in range(1,8):\n",
    "    A[1,i] = A[1,i-1]+A[0,i]\n",
    "print \"Median falls in the class (110−120) = l+((n/2−c)∗h)/f= \",\n",
    "Q2 = 110+(8*10)/50\n",
    "print Q2\n",
    "print \"Lower quartile also falls in the class (90−100)= \",\n",
    "Q1 = 90+(57.5-30)*10/35\n",
    "print Q1\n",
    "print \"Upper quartile also falls in the class (120−130)= \",\n",
    "Q3 = 120+(172.5-157)*10/45\n",
    "print Q3\n",
    "print \"Quartile coefficient of skewness= \",(Q1+Q3-2*Q2)/(Q3-Q1)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 23.13, page no. 631"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 76,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The first row of A denotes the corresponding I.R. of students \n",
      "The second row denotes the corresponding deviation of I.R.\n",
      "The third row denotes the square of corresponding deviation of I.R.\n",
      "The fourth row denotes the corresponding E.R. of students \n",
      "The fifth row denotes the corresponding deviation of E.R.\n",
      "The sixth row denotes the square of corresponding deviation of E.R.\n",
      "The seventh row denotes the product of the two corresponding deviations\n",
      "[[ 105.  104.  102.  101.  100.   99.   98.   96.   93.   92.]\n",
      " [   6.    5.    3.    2.    1.    0.   -1.   -3.   -6.   -7.]\n",
      " [  36.   25.    9.    4.    1.    0.    1.    9.   36.   49.]\n",
      " [ 101.  103.  100.   98.   95.   96.  104.   92.   97.   94.]\n",
      " [   3.    5.    2.    0.   -3.   -2.    6.   -6.   -1.   -4.]\n",
      " [   9.   25.    4.    0.    9.    4.   36.   36.    1.   16.]\n",
      " [  18.   25.    6.    0.   -3.   -0.   -6.   18.    6.   28.]]\n",
      "The sum of elements of first row=a =  990.0\n",
      "The sum of  elements of second row=b =  0.0\n",
      "The sum of elements of third row=c =  170.0\n",
      "The sum of elements of fourth row=d =  980.0\n",
      "The sum of elements of fifth row=e =  0.0\n",
      "The sum of elements of sixth row=d =  140.0\n",
      "The sum of elements of seventh row=d =  92.0\n",
      "Coefficient of correlation =  0.596347425668\n"
     ]
    }
   ],
   "source": [
    "import numpy\n",
    "\n",
    "A = numpy.zeros([7,10])\n",
    "print \"The first row of A denotes the corresponding I.R. of students \"\n",
    "A[0] = [105,104,102,101,100,99,98,96,93,92]\n",
    "print \"The second row denotes the corresponding deviation of I.R.\"\n",
    "for i in range(0,10):\n",
    "    A[1,i] = A[0,i]-99\n",
    "print \"The third row denotes the square of corresponding deviation of I.R.\"\n",
    "for i in range(0,10):\n",
    "    A[2,i] = A[1,i]**2\n",
    "print \"The fourth row denotes the corresponding E.R. of students \"\n",
    "A[3] = [101,103,100,98,95,96,104,92,97,94]\n",
    "print \"The fifth row denotes the corresponding deviation of E.R.\"\n",
    "for i in range(0,10):\n",
    "    A[4,i] = A[3,i]-98\n",
    "print \"The sixth row denotes the square of corresponding deviation of E.R.\"\n",
    "for i in range(0,10):\n",
    "    A[5,i] = A[4,i]**2\n",
    "print \"The seventh row denotes the product of the two corresponding deviations\"\n",
    "for i in range(0,10):\n",
    "    A[6,i] = A[1,i]*A[4,i]\n",
    "print A\n",
    "a = 0\n",
    "print \"The sum of elements of first row=a = \",\n",
    "for i in range(0,10):\n",
    "    a = a+A[0,i]\n",
    "print a\n",
    "b = 0 \n",
    "print \"The sum of  elements of second row=b = \",\n",
    "for i in range(0,10):\n",
    "    b = b+A[1,i]\n",
    "print b\n",
    "c = 0\n",
    "print \"The sum of elements of third row=c = \",\n",
    "for i in range(0,10):\n",
    "    c = c+A[2,i]\n",
    "print c\n",
    "d = 0\n",
    "print \"The sum of elements of fourth row=d = \",\n",
    "for i in range(0,10):\n",
    "    d = d+A[3,i]\n",
    "print d\n",
    "e = 0\n",
    "print \"The sum of elements of fifth row=e = \",\n",
    "for i in range(0,10):\n",
    "    e = e+A[4,i]\n",
    "print e\n",
    "f = 0\n",
    "print \"The sum of elements of sixth row=d = \",\n",
    "for i in range(0,10):\n",
    "    f = f+A[5,i]\n",
    "print f\n",
    "g = 0\n",
    "print \"The sum of elements of seventh row=d = \",\n",
    "for i in range(0,10):\n",
    "    g = g+A[6,i]\n",
    "print g\n",
    "print \"Coefficient of correlation = \",g/(c*f)**0.5"
   ]
  }
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