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NCERT Exemplar · Class 11 Mathematics Statistics

46 questions · 46 still being checked

EXERCISE 15.3 21–30 (part 3 of 5)

  1. Exercise 21

    Following are the marks obtained, out of 100\displaystyle 100 , by two students Ravi and Hashina in 10\displaystyle 10 tests.
    Ravi25\displaystyle 2550\displaystyle 5045\displaystyle 4530\displaystyle 3070\displaystyle 7042\displaystyle 4236\displaystyle 3648\displaystyle 4835\displaystyle 3560\displaystyle 60
    Hashina10\displaystyle 1070\displaystyle 7050\displaystyle 5020\displaystyle 2095\displaystyle 9555\displaystyle 5542\displaystyle 4260\displaystyle 6048\displaystyle 4880\displaystyle 80
    Who is more intelligent and who is more consistent?

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    Ravi:\[\bar{x}_R = \frac{441}{10} = 44.1, \qquad \sum x^2 = 21139 \]\[\sigma_R^2 = 2113.9 - (44.1)^2 = 169.09 \Rightarrow \sigma_R \approx 13.00 \]\[\text{C.V.}_R = \frac{\sigma_R}{\bar{x}_R}\times 100 = \frac{13.00}{44.1}\times 100 \approx 29.5\% \]Hashina:\[\bar{x}_H = \frac{530}{10} = 53, \qquad \sum x^2 = 34018 \]\[\sigma_H^2 = 3401.8 - 53^2 = 592.8 \Rightarrow \sigma_H \approx 24.35 \]\[\text{C.V.}_H = \frac{24.35}{53}\times 100 \approx 45.9\% \]\[\bar{x}_H > \bar{x}_R, \qquad \text{C.V.}_R < \text{C.V.}_H \]Answer: Hashina is better (higher mean marks); Ravi is more consistent (smaller C.V.).

  2. Exercise 22

    Mean and standard deviation of 100\displaystyle 100 observations were found to be 40\displaystyle 40 and 10\displaystyle 10, respectively. If at the time of calculation two observations were wrongly taken as 30\displaystyle 30 and 70\displaystyle 70 in place of 3\displaystyle 3 and 27\displaystyle 27 respectively, find the correct standard deviation.

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    NCERT’s answer
    10.$\displaystyle 24$
    \[\sum x = 100 \times 40 = 4000 \]\[\sum x^2 = 100\,(10^2 + 40^2) = 170000 \]Correct values:\[\sum x = 4000 - (30 + 70) + (3 + 27) = 3930 \Rightarrow \bar{x} = 39.3 \]\[\sum x^2 = 170000 - (30^2 + 70^2) + (3^2 + 27^2) = 170000 - 5800 + 738 = 164938 \]\[\sigma^2 = \frac{164938}{100} - (39.3)^2 = 1649.38 - 1544.49 = 104.89 \]\[\sigma = \sqrt{104.89} \approx 10.24 \]Answer: Correct standard deviation \(\displaystyle \approx 10.24\).

  3. Exercise 23

    While calculating the mean and variance of 10\displaystyle 10 readings, a student wrongly used the reading 52\displaystyle 52 for the correct reading 25\displaystyle 25 . He obtained the mean and variance as 45\displaystyle 45 and 16\displaystyle 16 respectively. Find the correct mean and the variance.

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    NCERT’s answer
    Mean \(\displaystyle =42.3\), Var. $\displaystyle 43.81$
    \[\sum x = 10 \times 45 = 450 \]\[\sum x^2 = 10\,(16 + 45^2) = 20410 \]Correct values:\[\sum x = 450 - 52 + 25 = 423 \Rightarrow \bar{x} = 42.3 \]\[\sum x^2 = 20410 - 52^2 + 25^2 = 20410 - 2704 + 625 = 18331 \]\[\sigma^2 = \frac{18331}{10} - (42.3)^2 = 1833.1 - 1789.29 = 43.81 \]Answer: Correct mean \(\displaystyle = 42.3\), correct variance \(\displaystyle = 43.81\).
  4. Choose the correct answer out of the given four options in each of the Exercises $\displaystyle 24$ to $\displaystyle 39$ (M.C.Q.).

    Exercise 24

    The mean deviation of the data 3\displaystyle 3, 10\displaystyle 10, 10\displaystyle 10, 4\displaystyle 4, 7\displaystyle 7, 10\displaystyle 10, 5\displaystyle 5 from the mean is
    (A)
    2\displaystyle 2 (B) 2.57\displaystyle 2.57
    (C)
    3\displaystyle 3 (D) 3.75\displaystyle 3.75

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    NCERT’s answer
    B
    (B) \(\displaystyle 2.57\)\[\bar{x} = \frac{3 + 10 + 10 + 4 + 7 + 10 + 5}{7} = \frac{49}{7} = 7 \]\[\text{M.D.} = \frac{4 + 3 + 3 + 3 + 0 + 3 + 2}{7} = \frac{18}{7} \approx 2.57 \]

  5. Exercise 25

    Mean deviation for n\displaystyle n observations x1,x2,…,xn\displaystyle x_1, x_2, \ldots, x_n from their mean xˉ\displaystyle \bar{x} is given by
    (A)
    ∑i=1n(xi−xˉ)\displaystyle \sum_{i=1}^n\left(x_i-\bar{x}\right)
    (B)
    1n∑i=1n∣xi−xˉ∣\displaystyle \frac{1}{n} \sum_{i=1}^n\left|x_i-\bar{x}\right|
    (C)
    ∑i=1n(xi−xˉ)2\displaystyle \sum_{i=1}^n\left(x_i-\bar{x}\right)^2
    (D)
    1n∑i=1n(xi−xˉ)2\displaystyle \frac{1}{n} \sum_{i=1}^n\left(x_i-\bar{x}\right)^2

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    NCERT’s answer
    B
    (B) \(\displaystyle \frac{1}{n}\sum_{i=1}^n \left|x_i-\bar{x}\right|\)The mean deviation averages the absolute deviations. Option (A) always equals \(\displaystyle 0\); (C) and (D) use squares, not absolute values.

  6. Exercise 26

    When tested, the lives (in hours) of 5\displaystyle 5 bulbs were noted as follows:
    1357\displaystyle 1357, 1090\displaystyle 1090, 1666\displaystyle 1666, 1494\displaystyle 1494, 1623\displaystyle 1623
    The mean deviations (in hours) from their mean is
    (A)
    178\displaystyle 178
    (B)
    179\displaystyle 179
    (C)
    220\displaystyle 220
    (D)
    356\displaystyle 356

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    (A) \(\displaystyle 178\)\[\bar{x} = \frac{1357+1090+1666+1494+1623}{5} = \frac{7230}{5} = 1446 \]\[|x_i-\bar{x}| = 89,\ 356,\ 220,\ 48,\ 177 \]\[\text{M.D.} = \frac{89+356+220+48+177}{5} = \frac{890}{5} = 178 \]

  7. Exercise 27

    Following are the marks obtained by 9\displaystyle 9 students in a mathematics test:
    50\displaystyle 50, 69\displaystyle 69, 20\displaystyle 20, 33\displaystyle 33, 53\displaystyle 53, 39\displaystyle 39, 40\displaystyle 40, 65\displaystyle 65, 59\displaystyle 59
    The mean deviation from the median is:
    (A)
    9\displaystyle 9 (B) 10.5\displaystyle 10.5
    (C)
    12.67\displaystyle 67
    (D)
    14.76\displaystyle 76

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    NCERT’s answer
    C
    (C) \(\displaystyle 12.67\)\[20,\ 33,\ 39,\ 40,\ 50,\ 53,\ 59,\ 65,\ 69 \quad \text{(in order)} \]\[M = 5\text{th value} = 50 \]\[|x_i-M| = 30,\ 17,\ 11,\ 10,\ 0,\ 3,\ 9,\ 15,\ 19 \]\[\text{M.D.} = \frac{114}{9} \approx 12.67 \]

  8. Exercise 28

    The standard deviation of the data 6\displaystyle 6, 5\displaystyle 5, 9\displaystyle 9, 13\displaystyle 13, 12\displaystyle 12, 8\displaystyle 8, 10\displaystyle 10 is
    (A)
    527\displaystyle \sqrt{\frac{52}{7}}
    (B)
    527\displaystyle \frac{52}{7}
    (C)
    6\displaystyle \sqrt{6}

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    NCERT’s answer
    A
    (A) \(\displaystyle \sqrt{\dfrac{52}{7}}\)\[\bar{x} = \frac{6+5+9+13+12+8+10}{7} = \frac{63}{7} = 9 \]\[\sum (x_i-\bar{x})^2 = 9+16+0+16+9+1+1 = 52 \]\[\sigma = \sqrt{\frac{52}{7}} \]

  9. Exercise 29

    Let x1,x2,…,xn\displaystyle x_1, x_2, \ldots, x_n be n\displaystyle n observations and xˉ\displaystyle \bar{x} be their arithmetic mean. The formula for the standard deviation is given by
    (A)
    ∑(xi−xˉ)2\displaystyle \sum\left(x_i-\bar{x}\right)^2
    (B)
    ∑(xi−xˉ)2n\displaystyle \frac{\sum\left(x_i-\bar{x}\right)^2}{n}
    (C)
    ∑(xi−xˉ)2n\displaystyle \sqrt{\frac{\sum\left(x_i-\bar{x}\right)^2}{n}}
    (D)
    ∑xi2n+xˉ2\displaystyle \sqrt{\frac{\sum x_i^2}{n}+\bar{x}^2}

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    NCERT’s answer
    C
    (C) \(\displaystyle \sqrt{\dfrac{\sum (x_i-\bar{x})^2}{n}}\)Standard deviation is the square root of the mean squared deviation. Option (D) has the wrong sign, since\[\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2 \]

  10. Exercise 30

    The mean of 100\displaystyle 100 observations is 50\displaystyle 50 and their standard deviation is 5. The sum of all squares of all the observations is
    (A)
    50000\displaystyle 50000
    (B)
    250000\displaystyle 250000
    (C)
    252500\displaystyle 252500
    (D)
    255000\displaystyle 255000

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    NCERT’s answer
    C
    (C) \(\displaystyle 252500\)\[\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2 \]\[25 = \frac{\sum x_i^2}{100} - 2500 \]\[\sum x_i^2 = 100 \times 2525 = 252500 \]