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NCERT Exemplar · Class 11 Mathematics Statistics

46 questions · 46 still being checked

EXERCISE 15.3 31–40 (part 4 of 5)

  1. Choose the correct answer out of the given four options in each of the Exercises $\displaystyle 24$ to $\displaystyle 39$ (M.C.Q.).

    Exercise 31

    Let a,b,c,d,e\displaystyle a, b, c, d, e be the observations with mean m\displaystyle m and standard deviation s\displaystyle s. The standard deviation of the observations a+k,b+k,c+k,d+k,e+k\displaystyle a+k, b+k, c+k, d+k, e+k is
    (A)
    s\displaystyle s (B) ks\displaystyle k s
    (C)
    s+k\displaystyle s+k
    (D)
    sk\displaystyle \frac{s}{k}

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    NCERT’s answer
    A
    (A) \(\displaystyle s\)Write the observations as \(\displaystyle x_i\), and the new ones as \(\displaystyle y_i = x_i + k\).\[\bar{y} = m + k \]\[y_i - \bar{y} = (x_i+k) - (m+k) = x_i - m \]\[\sigma_y = \sqrt{\frac{\sum (x_i-m)^2}{5}} = s \]

  2. Exercise 32

    Let x1,x2,x3,x4,x5\displaystyle x_1, x_2, x_3, x_4, x_5 be the observations with mean m\displaystyle m and standard deviation s\displaystyle s. The standard deviation of the observations kx1,kx2,kx3,kx4,kx5\displaystyle k x_1, k x_2, k x_3, k x_4, k x_5 is
    (A)
    k+s\displaystyle k+s
    (B)
    sk\displaystyle \frac{s}{k}
    (C)
    ks\displaystyle k s

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    NCERT’s answer
    C
    (C) \(\displaystyle ks\)Let \(\displaystyle y_i = kx_i\).\[\bar{y} = km \]\[y_i - \bar{y} = k(x_i - m) \]\[\sigma_y = \sqrt{\frac{k^2\sum (x_i-m)^2}{5}} = |k|\,s = ks \quad (k>0) \]

  3. Exercise 33

    Let x1,x2,…xn\displaystyle x_1, x_2, \ldots x_n be n\displaystyle n observations. Let wi=lxi+k\displaystyle w_i=l x_i+k for i=1,2,…n\displaystyle i=1,2, \ldots n, where l\displaystyle l and k\displaystyle k are constants. If the mean of xi\displaystyle x_i's is 48\displaystyle 48 and their standard deviation is 12\displaystyle 12, the mean of wi\displaystyle w_i's is 55\displaystyle 55 and standard deviation of wi\displaystyle w_i's is 15\displaystyle 15, the values of l\displaystyle l and k\displaystyle k should be
    (A)
    l=1.25,k=−5\displaystyle l=1.25, k=-5
    (B)
    l=−1.25,k=5\displaystyle l=-1.25, k=5
    (C)
    l=2.5,k=−5\displaystyle l=2.5, k=-5
    (D)
    l=2.5,k=5\displaystyle l=2.5, k=5

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    NCERT’s answer
    A
    (A) \(\displaystyle l=1.25,\ k=-5\). Standard deviation scales by \(\displaystyle |l|\); the mean satisfies \(\displaystyle \bar w = l\bar x + k\).\[\sigma_w = |l|\,\sigma_x \Rightarrow 15 = |l|(12) \Rightarrow |l| = 1.25 \]\[\bar w = l\bar x + k \Rightarrow 55 = 1.25(48) + k \]\[55 = 60 + k \Rightarrow k = -5 \]For \(\displaystyle l=-1.25\): \(\displaystyle k = 55 + 60 = 115\), which is not an option.Answer: (A)

  4. Exercise 34

    Standard deviations for first 10\displaystyle 10 natural numbers is
    (A)
    5.5\displaystyle 5
    (B)
    3.87\displaystyle 87
    (C)
    2.97\displaystyle 97
    (D)
    2.87\displaystyle 87

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    NCERT’s answer
    D
    (D) \(\displaystyle 2.87\).\[\sum x^2 = \frac{10\cdot 11\cdot 21}{6} = 385, \qquad \bar x = \frac{55}{10} = 5.5 \]\[\sigma^2 = \frac{385}{10} - (5.5)^2 = 38.5 - 30.25 = 8.25 \]\[\sigma = \sqrt{8.25} \approx 2.87 \]Answer: (D)

  5. Exercise 35

    Consider the numbers 1\displaystyle 1, 2\displaystyle 2, 3\displaystyle 3, 4\displaystyle 4, 5\displaystyle 5, 6\displaystyle 6, 7\displaystyle 7, 8\displaystyle 8, 9\displaystyle 9, 10. If 1\displaystyle 1 is added to each number, the variance of the numbers so obtained is
    (A)
    6.5\displaystyle 5
    (B)
    2.87\displaystyle 87
    (C)
    3.87\displaystyle 87
    (D)
    8.25\displaystyle 25

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    NCERT’s answer
    D
    (D) \(\displaystyle 8.25\). Adding a constant shifts the mean but leaves the spread unchanged.\[\operatorname{Var}(x+1) = \operatorname{Var}(x) \]\[\sigma^2 = \frac{385}{10} - (5.5)^2 = 8.25 \]Answer: (D)

  6. Exercise 36

    Consider the first 10\displaystyle 10 positive integers. If we multiply each number by -1\displaystyle 1 and then add 1\displaystyle 1 to each number, the variance of the numbers so obtained is
    (A)
    8.25\displaystyle 25
    (B)
    6.5\displaystyle 5
    (C)
    3.87\displaystyle 87
    (D)
    2.87\displaystyle 87

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    NCERT’s answer
    A
    (A) \(\displaystyle 8.25\). For \(\displaystyle y = ax + b\), \(\displaystyle \operatorname{Var}(y) = a^2\operatorname{Var}(x)\).\[y = -x + 1 \Rightarrow \operatorname{Var}(y) = (-1)^2\operatorname{Var}(x) \]\[\operatorname{Var}(x) = \frac{385}{10} - (5.5)^2 = 8.25 \]Answer: (A)

  7. Exercise 37

    The following information relates to a sample of size 60\displaystyle 60: ∑x2=18000\displaystyle \sum x^2=18000,
    ∑x=960\displaystyle \sum x=960
    The variance is
    (A)
    6.63\displaystyle 63
    (B)
    16\displaystyle 16
    (C)
    22\displaystyle 22
    (D)
    44\displaystyle 44

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    NCERT’s answer
    D
    (D) \(\displaystyle 44\).\[\sigma^2 = \frac{\sum x^2}{n} - \left(\frac{\sum x}{n}\right)^2 \]\[\sigma^2 = \frac{18000}{60} - \left(\frac{960}{60}\right)^2 \]\[\sigma^2 = 300 - 256 = 44 \]Answer: (D)

  8. Exercise 38

    Coefficient of variation of two distributions are 50\displaystyle 50 and 60\displaystyle 60, and their arithmetic means are 30\displaystyle 30 and 25\displaystyle 25 respectively. Difference of their standard deviation is
    (A)
    0\displaystyle 0 (B) 1\displaystyle 1 (C) 1.5\displaystyle 1.5
    (D)
    2.5\displaystyle 5

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    NCERT’s answer
    A
    (A) \(\displaystyle 0\).\[\text{C.V.} = \frac{\sigma}{\bar x}\times 100 \Rightarrow \sigma = \frac{\text{C.V.}\times\bar x}{100} \]\[\sigma_1 = \frac{50\times 30}{100} = 15 \]\[\sigma_2 = \frac{60\times 25}{100} = 15 \]\[\sigma_1 - \sigma_2 = 0 \]Answer: (A)

  9. Exercise 39

    The standard deviation of some temperature data in °C is 5. If the data were converted into ∘F\displaystyle { }^{\circ} \mathrm{F}, the variance would be
    (A)
    81\displaystyle 81
    (B)
    57\displaystyle 57
    (C)
    36\displaystyle 36
    (D)
    25\displaystyle 25

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    NCERT’s answer
    A
    (A) \(\displaystyle 81\). Fahrenheit and Celsius are related linearly, so the standard deviation scales by the slope.\[F = \frac{9}{5}C + 32 \Rightarrow \sigma_F = \frac{9}{5}\sigma_C = \frac{9}{5}(5) = 9 \]\[\sigma_F^2 = 9^2 = 81 \]Answer: (A)
  10. Fill in the blanks in Exercises from $\displaystyle 40$ to 46.

    Exercise 40

    Coefficient of variation =⋯ Mean ×100\displaystyle =\frac{\cdots}{\text { Mean }} \times 100

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    NCERT’s answer
    SD
    Standard deviation\[\text{Coefficient of variation} = \frac{\text{Standard deviation}}{\text{Mean}} \times 100 \]Answer: Standard deviation