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NCERT Exemplar · Class 11 Mathematics Statistics

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EXERCISE 15.3 1–10 (part 1 of 5)

  1. Exercise 1

    Find the mean deviation about the mean of the distribution:
    Size20\displaystyle 2021\displaystyle 2122\displaystyle 2223\displaystyle 2324\displaystyle 24
    Frequency6\displaystyle 64\displaystyle 45\displaystyle 51\displaystyle 14\displaystyle 4

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    Mean first.\[\sum f_i x_i = 120+84+110+23+96 = 433,\qquad N = 20 \]\[\bar{x} = \frac{433}{20} = 21.65 \]\[\sum f_i\,|x_i-\bar{x}| = 6(1.65)+4(0.65)+5(0.35)+1(1.35)+4(2.35) \]\[= 9.9+2.6+1.75+1.35+9.4 = 25 \]\[\text{M.D.} = \frac{25}{20} = 1.25 \]Answer: \(\displaystyle 1.25\)

  2. Exercise 2

    Find the mean deviation about the median of the following distribution:
    Marks obtained10\displaystyle 1011\displaystyle 1112\displaystyle 1214\displaystyle 1415\displaystyle 15
    No. of students2\displaystyle 23\displaystyle 38\displaystyle 83\displaystyle 34\displaystyle 4

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    NCERT’s answer
    1.$\displaystyle 25$
    \[N = 2+3+8+3+4 = 20 \]Cumulative frequencies are \(\displaystyle 2,\,5,\,13,\,16,\,20\), so the 10th and 11th terms are both \(\displaystyle 12\).\[\text{Median} = \frac{12+12}{2} = 12 \]\[\sum f_i\,|x_i-12| = 2(2)+3(1)+8(0)+3(2)+4(3) \]\[= 4+3+0+6+12 = 25 \]\[\text{M.D.} = \frac{25}{20} = 1.25 \]Answer: \(\displaystyle 1.25\)

  3. Exercise 3

    Calculate the mean deviation about the mean of the set of first n\displaystyle n natural numbers when n\displaystyle n is an odd number.

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    NCERT’s answer
    \(\displaystyle \frac{n^2-1}{4 n}\)
    Take \(\displaystyle n = 2m+1\).\[\bar{x} = \frac{1+2+\cdots+n}{n} = \frac{n+1}{2} = m+1 \]The deviations \(\displaystyle |k-\bar{x}|\) are \(\displaystyle m,\,m-1,\,\ldots,\,1,\,0,\,1,\,\ldots,\,m\).\[\sum_{k=1}^{n}|k-\bar{x}| = 2\,(1+2+\cdots+m) = m(m+1) \]\[m(m+1) = \frac{n-1}{2}\cdot\frac{n+1}{2} = \frac{n^2-1}{4} \]\[\text{M.D.} = \frac{1}{n}\cdot\frac{n^2-1}{4} \]Answer: \(\displaystyle \dfrac{n^2-1}{4n}\)

  4. Exercise 4

    Calculate the mean deviation about the mean of the set of first n\displaystyle n natural numbers when n\displaystyle n is an even number.

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    NCERT’s answer
    \(\displaystyle \frac{n}{4}\)
    Take \(\displaystyle n = 2m\).\[\bar{x} = \frac{n+1}{2} = m+\tfrac{1}{2} \]The deviations are \(\displaystyle m-\tfrac12,\,\ldots,\,\tfrac32,\,\tfrac12\) for \(\displaystyle k\le m\), and \(\displaystyle \tfrac12,\,\tfrac32,\,\ldots,\,m-\tfrac12\) for \(\displaystyle k>m\).\[\sum_{k=1}^{n}|k-\bar{x}| = 2\left[\frac{1+3+5+\cdots+(2m-1)}{2}\right] \]\[= 1+3+5+\cdots+(2m-1) = m^2 \]\[\text{M.D.} = \frac{m^2}{2m} = \frac{m}{2} = \frac{n}{4} \]Answer: \(\displaystyle \dfrac{n}{4}\)

  5. Exercise 5

    Find the standard deviation of the first n\displaystyle n natural numbers.

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    NCERT’s answer
    \(\displaystyle \sqrt{\frac{n^2-1}{12}}\)
    \[\bar{x} = \frac{n+1}{2},\qquad \sum x_i^2 = \frac{n(n+1)(2n+1)}{6} \]\[\sigma^2 = \frac{1}{n}\sum x_i^2-\bar{x}^{\,2} = \frac{(n+1)(2n+1)}{6}-\frac{(n+1)^2}{4} \]\[= \frac{(n+1)\left[\,2(2n+1)-3(n+1)\,\right]}{12} = \frac{(n+1)(n-1)}{12} \]\[\sigma = \sqrt{\frac{n^2-1}{12}} \]Answer: \(\displaystyle \sqrt{\dfrac{n^2-1}{12}}\)

  6. Exercise 6

    The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results: Number of observations =25\displaystyle =25, mean =18.2\displaystyle =18.2 seconds, standard deviation =3.25\displaystyle =3.25 seconds. Further, another set of 15\displaystyle 15 observations x1,x2,…,x15\displaystyle x_1, x_2, \ldots, x_{15}, also in seconds, is now available and we have ∑i=115xi=279\displaystyle \sum_{i=1}^{15} x_i=279 and ∑i=115xi2=5524\displaystyle \sum_{i=1}^{15} x_i^2=5524. Calculate the standard derivation based on all 40\displaystyle 40 observations.

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    NCERT’s answer
    3.$\displaystyle 87$
    First $\displaystyle 25$ observations, using \(\displaystyle \sigma^2 = \dfrac{\sum x^2}{n}-\bar{x}^{\,2}\):\[\sum x = 25(18.2) = 455 \]\[\frac{\sum x^2}{25} = (3.25)^2+(18.2)^2 = 10.5625+331.24 = 341.8025 \]\[\sum x^2 = 8545.0625 \]All $\displaystyle 40$ observations:\[\sum x = 455+279 = 734,\qquad \sum x^2 = 8545.0625+5524 = 14069.0625 \]\[\bar{x} = \frac{734}{40} = 18.35 \]\[\sigma^2 = \frac{14069.0625}{40}-(18.35)^2 = 351.7266-336.7225 = 15.0041 \]\[\sigma = \sqrt{15.0041} \approx 3.87 \]Answer: \(\displaystyle \approx 3.87\) seconds

  7. Exercise 7

    The mean and standard deviation of a set of n1\displaystyle n_1 observations are xˉ1\displaystyle \bar{x}_1 and s1\displaystyle s_1, respectively while the mean and standard deviation of another set of n2\displaystyle n_2 observations are xˉ2\displaystyle \bar{x}_2 and s2\displaystyle s_2, respectively. Show that the standard deviation of the combined set of (n1+n2)\displaystyle \left(n_1+n_2\right) observations is given by S.D.=n1(s1)2+n2(s2)2n1+n2+n1n2(xˉ1−xˉ2)2(n1+n2)2\text{S.D.}=\sqrt{\frac{n_1\left(s_1\right)^2+n_2\left(s_2\right)^2}{n_1+n_2}+\frac{n_1 n_2\left(\bar{x}_1-\bar{x}_2\right)^2}{\left(n_1+n_2\right)^2}}

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    NCERT’s answer
    \(\displaystyle \sqrt{\frac{n_1\left(s_1\right)^2+n_2\left(s_2\right)^2}{n_1+n_2}+\frac{n_1 n_2\left(\bar{x}_1-\bar{x}_2\right)^2}{\left(n_1+n_2\right)^2}}\)
    Let \(\displaystyle N = n_1+n_2\), \(\displaystyle \Delta = \bar{x}_1-\bar{x}_2\), and let \(\displaystyle \bar{x}\) be the combined mean.\[\bar{x} = \frac{n_1\bar{x}_1+n_2\bar{x}_2}{N} \]\[d_1 = \bar{x}_1-\bar{x} = \frac{n_2\Delta}{N},\qquad d_2 = \bar{x}_2-\bar{x} = -\frac{n_1\Delta}{N} \]Write \(\displaystyle x_i-\bar{x} = (x_i-\bar{x}_1)+d_1\); the cross term vanishes because \(\displaystyle \sum (x_i-\bar{x}_1)=0\).\[\sum_{i=1}^{n_1}(x_i-\bar{x})^2 = \sum_{i=1}^{n_1}(x_i-\bar{x}_1)^2+n_1d_1^2 = n_1s_1^2+n_1d_1^2 \]\[\sum_{j=1}^{n_2}(y_j-\bar{x})^2 = n_2s_2^2+n_2d_2^2 \]\[\text{S.D.}^2 = \frac{n_1s_1^2+n_2s_2^2+n_1d_1^2+n_2d_2^2}{N} \]\[n_1d_1^2+n_2d_2^2 = \frac{n_1n_2^2+n_2n_1^2}{N^2}\,\Delta^2 = \frac{n_1n_2\,\Delta^2}{N} \]\[\text{S.D.}^2 = \frac{n_1s_1^2+n_2s_2^2}{n_1+n_2}+\frac{n_1n_2(\bar{x}_1-\bar{x}_2)^2}{(n_1+n_2)^2} \]Taking the square root gives the required result.

  8. Exercise 8

    Two sets each of 20\displaystyle 20 observations, have the same standard derivation 5. The first set has a mean 17\displaystyle 17 and the second a mean 22. Determine the standard deviation of the set obtained by combining the given two sets.

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    NCERT’s answer
    5.$\displaystyle 59$
    \[\bar{x} = \frac{20(17)+20(22)}{40} = 19.5 \]\[d_1 = 17-19.5 = -2.5,\qquad d_2 = 22-19.5 = 2.5 \]For each set, \(\displaystyle \sum (x-\bar{x})^2 = n\,s^2+n\,d^2\).\[\sigma^2 = \frac{20\left(25+6.25\right)+20\left(25+6.25\right)}{40} = 31.25 \]\[\sigma = \sqrt{31.25} = \frac{5\sqrt{5}}{2} \approx 5.59 \]Answer: \(\displaystyle \dfrac{5\sqrt{5}}{2}\approx 5.59\)

  9. Exercise 9

    The frequency distribution:
    x\displaystyle \boldsymbol{x}A2A3A4A5A6A
    f\displaystyle \boldsymbol{f}2\displaystyle 21\displaystyle 11\displaystyle 11\displaystyle 11\displaystyle 11\displaystyle 1
    where A is a positive integer, has a variance of 160. Determine the value of A.

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    NCERT’s answer
    $\displaystyle 7$
    \[N = \sum f = 2+1+1+1+1+1 = 7 \] \[\sum fx = 2A + 2A + 3A + 4A + 5A + 6A = 22A \] \[\sum fx^2 = 2A^2 + 4A^2 + 9A^2 + 16A^2 + 25A^2 + 36A^2 = 92A^2 \] \[\sigma^2 = \frac{\sum fx^2}{N} - \left(\frac{\sum fx}{N}\right)^2 = \frac{92A^2}{7} - \frac{484A^2}{49} = \frac{160A^2}{49} \] \[\frac{160A^2}{49} = 160 \;\Rightarrow\; A^2 = 49 \] \[A > 0 \;\Rightarrow\; A = 7 \] Answer: \(\displaystyle A = 7\)

  10. Exercise 10

    For the frequency distribution:
    x\displaystyle \boldsymbol{x}2\displaystyle 23\displaystyle 34\displaystyle 45\displaystyle 56\displaystyle 67\displaystyle 7
    f\displaystyle f4\displaystyle 49\displaystyle 916\displaystyle 1614\displaystyle 1411\displaystyle 116\displaystyle 6
    Find the standard distribution.

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    NCERT’s answer
    1.$\displaystyle 38$
    \[N = 4+9+16+14+11+6 = 60 \] \[\sum fx = 8+27+64+70+66+42 = 277 \] \[\sum fx^2 = 16+81+256+350+396+294 = 1393 \] \[\sigma^2 = \frac{1393}{60} - \left(\frac{277}{60}\right)^2 = \frac{83580 - 76729}{3600} = \frac{6851}{3600} \] \[\sigma = \frac{\sqrt{6851}}{60} \approx 1.38 \] Answer: standard deviation \(\displaystyle \approx 1.38\)