Added(A)/Deleted(D) following books

 A  Engineering_Thermodynamics_by_Dr._S._S._Khandare/README.txt
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter1.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter10.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter13.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter2.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter21.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter22.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter23.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter24.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter26.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter27.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter28.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter34.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter35.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter4.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter5.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter6.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter9.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/screenshots/screen1.png
A  Higher_Engineering_Mathematics_by_B._S._Grewal/screenshots/screen2.png
A  Higher_Engineering_Mathematics_by_B._S._Grewal/screenshots/screen3.png
A  sample_notebooks/ManikandanD/Chapter_2_Light_propagation_in_optical_fiber.ipynb

1 files changed, 762 insertions, 0 deletions

diff --git a/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter23.ipynb b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter23.ipynb
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+{
+ "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"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}