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NCERT Exemplar · Class 11 Mathematics Statistics

46 questions · 46 still being checked

EXERCISE 15.3 11–20 (part 2 of 5)

  1. Exercise 11

    There are 60\displaystyle 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test:
    Marks0\displaystyle 01\displaystyle 12\displaystyle 23\displaystyle 34\displaystyle 45\displaystyle 5
    Frequencyx−2\displaystyle x-2x\displaystyle xx2\displaystyle x^2(x+1)2\displaystyle (x+1)^22x\displaystyle 2 xx+1\displaystyle x+1
    where x\displaystyle x is a positive integer. Determine the mean and standard deviation of the marks.

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    NCERT’s answer
    Mean \(\displaystyle =2.8, \mathrm{SD}=1.12\)
    \[(x-2) + x + x^2 + (x+1)^2 + 2x + (x+1) = 60 \] \[2x^2 + 7x - 60 = 0 \] \[(2x+15)(x-4) = 0 \] \[x > 0 \;\Rightarrow\; x = 4 \] Frequencies for marks \(\displaystyle 0,1,2,3,4,5\) are \(\displaystyle 2,4,16,25,8,5\). \[\sum fx = 0+4+32+75+32+25 = 168 \] \[\bar{x} = \frac{168}{60} = 2.8 \] \[\sum fx^2 = 0+4+64+225+128+125 = 546 \] \[\sigma^2 = \frac{546}{60} - (2.8)^2 = 9.1 - 7.84 = 1.26 \] \[\sigma = \sqrt{1.26} \approx 1.12 \] Answer: mean \(\displaystyle = 2.8\), standard deviation \(\displaystyle \approx 1.12\)

  2. Exercise 12

    The mean life of a sample of 60\displaystyle 60 bulbs was 650\displaystyle 650 hours and the standard deviation was 8\displaystyle 8 hours. A second sample of 80\displaystyle 80 bulbs has a mean life of 660\displaystyle 660 hours and standard deviation 7\displaystyle 7 hours. Find the overall standard deviation.

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    NCERT’s answer
    8.$\displaystyle 9$
    \[\bar{x} = \frac{60(650) + 80(660)}{140} = \frac{91800}{140} = \frac{4590}{7} \] \[d_1 = 650 - \frac{4590}{7} = -\frac{40}{7}, \qquad d_2 = 660 - \frac{4590}{7} = \frac{30}{7} \] \[\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1+n_2} \] \[\sigma^2 = \frac{60\left(64 + \dfrac{1600}{49}\right) + 80\left(49 + \dfrac{900}{49}\right)}{140} = \frac{3916}{49} \] \[\sigma = \frac{\sqrt{3916}}{7} \approx 8.94 \] Answer: overall standard deviation \(\displaystyle \approx 8.94\) hours

  3. Exercise 13

    Mean and standard deviation of 100\displaystyle 100 items are 50\displaystyle 50 and 4\displaystyle 4, respectively. Find the sum of all the item and the sum of the squares of the items.

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    NCERT’s answer
    $\displaystyle 5000$, $\displaystyle 251600$
    \[\sum x = n\bar{x} = 100 \times 50 = 5000 \] \[\sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2 \;\Rightarrow\; 16 = \frac{\sum x^2}{100} - 2500 \] \[\sum x^2 = 100 \times 2516 = 251600 \] Answer: \(\displaystyle \sum x = 5000\), \(\displaystyle \sum x^2 = 251600\)

  4. Exercise 14

    If for a distribution ∑(x−5)=3,∑(x−5)2=43\displaystyle \sum(x-5)=3, \sum(x-5)^2=43 and the total number of item is 18\displaystyle 18 , find the mean and standard deviation.

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    NCERT’s answer
    Mean \(\displaystyle =5.17, \mathrm{SD}=1.53\)
    Let \(\displaystyle d = x - 5\), so \(\displaystyle n = 18\), \(\displaystyle \sum d = 3\), \(\displaystyle \sum d^2 = 43\). \[\bar{d} = \frac{3}{18} = \frac{1}{6} \] \[\bar{x} = 5 + \bar{d} = \frac{31}{6} \approx 5.17 \] \[\sigma_x^2 = \sigma_d^2 = \frac{43}{18} - \left(\frac{1}{6}\right)^2 = \frac{85}{36} \] \[\sigma = \frac{\sqrt{85}}{6} \approx 1.54 \] Answer: mean \(\displaystyle = \dfrac{31}{6}\), standard deviation \(\displaystyle = \dfrac{\sqrt{85}}{6}\)

  5. Exercise 15

    Find the mean and variance of the frequency distribution given below:
    x\displaystyle \boldsymbol{x}1≤x<3\displaystyle 1 \leq x<33≤x<5\displaystyle 3 \leq x<55≤x<7\displaystyle 5 \leq x<77≤x<10\displaystyle 7 \leq x<10
    f\displaystyle f6\displaystyle 64\displaystyle 45\displaystyle 51\displaystyle 1

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    Class marks \(\displaystyle x_i = 2,\,4,\,6,\,8.5\). \[N = 6+4+5+1 = 16 \] \[\sum fx = 12+16+30+8.5 = 66.5 \] \[\bar{x} = \frac{66.5}{16} = \frac{133}{32} \approx 4.16 \] \[\sum fx^2 = 24+64+180+72.25 = 340.25 \] \[\sigma^2 = \frac{340.25}{16} - \left(\frac{66.5}{16}\right)^2 = \frac{4087}{1024} \approx 3.99 \] Answer: mean \(\displaystyle \approx 4.16\), variance \(\displaystyle \approx 3.99\)

  6. Exercise 16

    Calculate the mean deviation about the mean for the following frequency distribution:
    Class interval0\displaystyle 0-4\displaystyle 44\displaystyle 4-8\displaystyle 88\displaystyle 8-12\displaystyle 1212\displaystyle 12-16\displaystyle 1616\displaystyle 16-20\displaystyle 20
    Frequency4\displaystyle 46\displaystyle 68\displaystyle 85\displaystyle 52\displaystyle 2

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    Class marks \(\displaystyle x_i = 2,\,6,\,10,\,14,\,18\), \(\displaystyle N = 25\). \[\sum fx = 8+36+80+70+36 = 230 \] \[\bar{x} = \frac{230}{25} = 9.2 \] \[|x_i - \bar{x}| = 7.2,\; 3.2,\; 0.8,\; 4.8,\; 8.8 \] \[\sum f|x_i - \bar{x}| = 28.8 + 19.2 + 6.4 + 24 + 17.6 = 96 \] \[\text{M.D.} = \frac{96}{25} = 3.84 \] Answer: mean deviation about the mean \(\displaystyle = 3.84\)

  7. Exercise 17

    Calculate the mean deviation from the median of the following data:
    Class interval0\displaystyle 0-6\displaystyle 66\displaystyle 6-12\displaystyle 1212\displaystyle 12-18\displaystyle 1818\displaystyle 18-24\displaystyle 2424\displaystyle 24-30\displaystyle 30
    Frequency4\displaystyle 45\displaystyle 53\displaystyle 36\displaystyle 62\displaystyle 2

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    Cumulative frequencies:\[\begin{array}{c|ccccc} \text{Class} & 0\text{–}6 & 6\text{–}12 & 12\text{–}18 & 18\text{–}24 & 24\text{–}30 \\ \hline f & 4 & 5 & 3 & 6 & 2 \\ cf & 4 & 9 & 12 & 18 & 20 \end{array} \]\[N = 20, \quad \tfrac{N}{2} = 10 \Rightarrow \text{median class } 12\text{–}18 \]\[\text{Median} = l + \frac{\dfrac{N}{2} - cf}{f}\,h = 12 + \frac{10 - 9}{3}\times 6 = 14 \]\[\begin{array}{c|ccccc} x_i & 3 & 9 & 15 & 21 & 27 \\ \hline |x_i - 14| & 11 & 5 & 1 & 7 & 13 \\ f_i\,|x_i - 14| & 44 & 25 & 3 & 42 & 26 \end{array} \]\[\sum f_i\,|x_i - 14| = 140 \]\[\text{M.D.} = \frac{\sum f_i\,|x_i - 14|}{N} = \frac{140}{20} = 7 \]Answer: Mean deviation from the median \(\displaystyle = 7\).

  8. Exercise 18

    Determine the mean and standard deviation for the following distribution:
    Marks2\displaystyle 23\displaystyle 34\displaystyle 45\displaystyle 56\displaystyle 67\displaystyle 78\displaystyle 89\displaystyle 910\displaystyle 1011\displaystyle 1112\displaystyle 1213\displaystyle 1314\displaystyle 1415\displaystyle 1516\displaystyle 16
    Frequency1\displaystyle 16\displaystyle 66\displaystyle 68\displaystyle 88\displaystyle 82\displaystyle 22\displaystyle 23\displaystyle 30\displaystyle 02\displaystyle 21\displaystyle 10\displaystyle 00\displaystyle 00\displaystyle 01\displaystyle 1

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    NCERT’s answer
    Mean \(\displaystyle =\frac{239}{40}, \mathrm{SD}=2.85\)
    \[N = \sum f = 40 \]\[\sum fx = 2 + 18 + 24 + 40 + 48 + 14 + 16 + 27 + 22 + 12 + 16 = 239 \]\[\sum fx^2 = 4 + 54 + 96 + 200 + 288 + 98 + 128 + 243 + 242 + 144 + 256 = 1753 \]\[\bar{x} = \frac{239}{40} = 5.975 \]\[\sigma^2 = \frac{\sum fx^2}{N} - \bar{x}^2 = \frac{1753}{40} - (5.975)^2 = 43.825 - 35.7006 = 8.1244 \]\[\sigma = \sqrt{8.1244} \approx 2.85 \]Answer: Mean \(\displaystyle = 5.975\), standard deviation \(\displaystyle \approx 2.85\).

  9. Exercise 19

    The weights of coffee in 70\displaystyle 70 jars is shown in the following table:
    Weight (in grams)Frequency
    200\displaystyle 200-201\displaystyle 20113\displaystyle 13
    201\displaystyle 201-202\displaystyle 20227\displaystyle 27
    202\displaystyle 202-203\displaystyle 20318\displaystyle 18
    203\displaystyle 203-204\displaystyle 20410\displaystyle 10
    204\displaystyle 204-205\displaystyle 2051\displaystyle 1
    205\displaystyle 205-206\displaystyle 2061\displaystyle 1
    Determine variance and standard deviation of the above distribution.

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    NCERT’s answer
    Var. \(\displaystyle =1.16 \mathrm{gm}, \mathrm{S} . \mathrm{D}=1.08 \mathrm{gm}\)
    Take class marks \(\displaystyle x_i\) and shift by \(\displaystyle A = 202.5\), \(\displaystyle d_i = x_i - A\) (a shift leaves the variance unchanged).\[\begin{array}{c|cccccc} x_i & 200.5 & 201.5 & 202.5 & 203.5 & 204.5 & 205.5 \\ \hline f_i & 13 & 27 & 18 & 10 & 1 & 1 \\ d_i & -2 & -1 & 0 & 1 & 2 & 3 \\ f_i d_i & -26 & -27 & 0 & 10 & 2 & 3 \\ f_i d_i^2 & 52 & 27 & 0 & 10 & 4 & 9 \end{array} \]\[N = 70, \quad \sum f_i d_i = -38, \quad \sum f_i d_i^2 = 102 \]\[\sigma^2 = \frac{\sum f_i d_i^2}{N} - \left(\frac{\sum f_i d_i}{N}\right)^2 = \frac{102}{70} - \left(\frac{38}{70}\right)^2 = \frac{5696}{4900} \approx 1.16 \]\[\sigma = \sqrt{1.1624} \approx 1.08 \]Answer: Variance \(\displaystyle \approx 1.16\), standard deviation \(\displaystyle \approx 1.08\) g.

  10. Exercise 20

    Determine mean and standard deviation of first n\displaystyle n terms of an A.P. whose first term is a\displaystyle a and common difference is d\displaystyle d.

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    NCERT’s answer
    Mean \(\displaystyle =a+\frac{d(n-1)}{2}\), S.D \(\displaystyle =d \sqrt{\frac{n^2-1}{12}}\)
    \[x_k = a + (k-1)d, \quad k = 1, 2, \dots, n \]\[\bar{x} = \frac{1}{n}\cdot\frac{n}{2}\,[\,2a + (n-1)d\,] = a + \frac{(n-1)d}{2} \]\[x_k - \bar{x} = \left(k - \frac{n+1}{2}\right)d \]\[\sigma^2 = \frac{d^2}{n}\sum_{k=1}^{n}\left(k - \frac{n+1}{2}\right)^2 \]\[\sum_{k=1}^{n}\left(k - \frac{n+1}{2}\right)^2 = \sum k^2 - \frac{n(n+1)^2}{4} = \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)^2}{4} = \frac{n(n^2-1)}{12} \]\[\sigma^2 = \frac{d^2\,(n^2-1)}{12} \]Answer: Mean \(\displaystyle = a + \dfrac{(n-1)d}{2}\), standard deviation \(\displaystyle = |d|\sqrt{\dfrac{n^2-1}{12}}\).