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NCERT Exemplar · Class 9 Mathematics Statistics and Probability

72 questions · 72 still being checked

EXERCISE 14.1 11–20 (part 2 of 7)

  1. The number of classes in the distribution will be :

    Exercise 11

    The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is : (A) 28\displaystyle 28 (B) 30\displaystyle 30 (C) 35\displaystyle 35 (D) 38\displaystyle 38

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 38\) \[\text{Sum of 5 numbers} = 5\times 30 = 150 \] \[\text{Sum of remaining 4} = 4\times 28 = 112 \] \[\text{Excluded number} = 150-112 = 38 \]

  2. Exercise 12

    If the mean of the observations: x,x+3,x+5,x+7,x+10\displaystyle x, x+3, x+5, x+7, x+10 is 9\displaystyle 9, the mean of the last three observations is (A) 1013\displaystyle 10 \frac{1}{3} (B) 1023\displaystyle 10 \frac{2}{3} (C) 1113\displaystyle 11 \frac{1}{3} (D) 1123\displaystyle 11 \frac{2}{3}

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 11\tfrac{1}{3}\) \[\frac{x+(x+3)+(x+5)+(x+7)+(x+10)}{5} = 9 \] \[5x+25=45 \quad\Rightarrow\quad x=4 \] \[\text{Mean of last three} = \frac{(x+5)+(x+7)+(x+10)}{3} = \frac{3x+22}{3} = \frac{34}{3} = 11\tfrac{1}{3} \]

  3. Exercise 13

    If xˉ\displaystyle \bar{x} represents the mean of n\displaystyle n observations x1,x2,…,xn\displaystyle x_{1}, x_{2}, \ldots, x_{n}, then value of ∑i=1n(xi−xˉ)\displaystyle \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right) is: (A) -1\displaystyle 1 (B) 0\displaystyle 0 (C) 1\displaystyle 1 (D) n−1\displaystyle n-1

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 0\). \[\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i \quad\Rightarrow\quad \sum_{i=1}^{n} x_i = n\bar{x} \] \[\sum_{i=1}^{n}(x_i-\bar{x}) = \sum_{i=1}^{n} x_i - n\bar{x} = n\bar{x}-n\bar{x} = 0 \]

  4. Exercise 14

    If each observation of the data is increased by 5\displaystyle 5, then their mean (A) remains the same (B) becomes 5\displaystyle 5 times the original mean (C) is decreased by 5\displaystyle 5 (D) is increased by 5\displaystyle 5

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    NCERT’s answer
    (D)
    (D) is increased by 5. \[\bar{x}_{\text{new}} = \frac{\sum_{i=1}^{n}(x_i+5)}{n} = \frac{\sum_{i=1}^{n} x_i}{n} + 5 = \bar{x}+5 \]

  5. Exercise 15

    Let xˉ\displaystyle \bar{x} be the mean of x1,x2,…,xn\displaystyle x_{1}, x_{2}, \ldots, x_{n} and yˉ\displaystyle \bar{y} the mean of y1,y2,…,yn\displaystyle y_{1}, y_{2}, \ldots, y_{n}. If zˉ\displaystyle \bar{z} is the mean of x1,x2,…,xn,y1,y2,…,yn\displaystyle x_{1}, x_{2}, \ldots, x_{n}, y_{1}, y_{2}, \ldots, y_{n}, then zˉ\displaystyle \bar{z} is equal to (A) xˉ+yˉ\displaystyle \bar{x}+\bar{y} (B) xˉ+yˉ2\displaystyle \frac{\bar{x}+\bar{y}}{2} (C) xˉ+yˉn\displaystyle \frac{\bar{x}+\bar{y}}{n} (D) xˉ+yˉ2n\displaystyle \frac{\bar{x}+\bar{y}}{2 n}

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    NCERT’s answer
    (B)
    (B) \(\displaystyle \dfrac{\bar{x}+\bar{y}}{2}\). \[\bar{x}=\frac{\sum_{i=1}^{n} x_i}{n},\qquad \bar{y}=\frac{\sum_{i=1}^{n} y_i}{n} \] \[\bar{z} = \frac{\sum_{i=1}^{n} x_i+\sum_{i=1}^{n} y_i}{2n} = \frac{n\bar{x}+n\bar{y}}{2n} = \frac{\bar{x}+\bar{y}}{2} \]

  6. Exercise 16

    If xˉ\displaystyle \bar{x} is the mean of x1,x2,…,xn\displaystyle x_{1}, x_{2}, \ldots, x_{n}, then for a≠0\displaystyle a \neq 0, the mean of ax1,ax2,…,axn,x1a\displaystyle a x_{1}, a x_{2}, \ldots, a x_{n}, \frac{x_{1}}{a}, x2a,…,xna\displaystyle \frac{x_{2}}{a}, \ldots, \frac{x_{n}}{a} is (A) (a+1a)xˉ\displaystyle \left(a+\frac{1}{a}\right) \bar{x} (B) (a+1a)xˉ2\displaystyle \left(a+\frac{1}{a}\right) \frac{\bar{x}}{2} (C) (a+1a)xˉn\displaystyle \left(a+\frac{1}{a}\right) \frac{\bar{x}}{n} (D) (a+1a)xˉ2n\displaystyle \frac{\left(a+\dfrac{1}{a}\right) \bar{x}}{2 n}

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    NCERT’s answer
    (B)
    (B) \(\displaystyle \left(a+\dfrac{1}{a}\right)\dfrac{\bar{x}}{2}\). \[\sum_{i=1}^{n} ax_i + \sum_{i=1}^{n}\frac{x_i}{a} = \left(a+\frac{1}{a}\right)\sum_{i=1}^{n} x_i = \left(a+\frac{1}{a}\right)n\bar{x} \] \[\text{Mean} = \frac{\left(a+\frac1a\right)n\bar{x}}{2n} = \left(a+\frac{1}{a}\right)\frac{\bar{x}}{2} \]

  7. Exercise 17

    If xˉ1,xˉ2,xˉ3,…,xˉn\displaystyle \bar{x}_{1}, \bar{x}_{2}, \bar{x}_{3}, \ldots, \bar{x}_{n} are the means of n\displaystyle n groups with n1,n2,…,nn\displaystyle n_{1}, n_{2}, \ldots, n_{n} number of observations respectively, then the mean xˉ\displaystyle \bar{x} of all the groups taken together is given by : (A) ∑i=1nnixˉi\displaystyle \sum_{i=1}^{n} n_{i} \bar{x}_{i} (B) ∑i=1nnixˉin2\displaystyle \frac{\sum_{i=1}^{n} n_{i} \bar{x}_{i}}{n^{2}} (C) ∑i=1nnixˉi∑i=1nni\displaystyle \frac{\sum_{i=1}^{n} n_{i} \bar{x}_{i}}{\sum_{i=1}^{n} n_{i}} (D) ∑i=1nnixˉi2n\displaystyle \frac{\sum_{i=1}^{n} n_{i} \bar{x}_{i}}{2 n}

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    NCERT’s answer
    (C)
    (C) \(\displaystyle \dfrac{\sum_{i=1}^{n} n_i\bar{x}_i}{\sum_{i=1}^{n} n_i}\). \[\text{Total sum} = \sum_{i=1}^{n} n_i\bar{x}_i, \qquad \text{Total count} = \sum_{i=1}^{n} n_i \] \[\bar{x} = \frac{\sum_{i=1}^{n} n_i\bar{x}_i}{\sum_{i=1}^{n} n_i} \]

  8. Exercise 18

    The mean of 100\displaystyle 100 observations is 50\displaystyle 50 . If one of the observations which was 50\displaystyle 50 is replaced by 150\displaystyle 150, the resulting mean will be : (A) 50.5\displaystyle 50.5 (B) 51\displaystyle 51 (C) 51.5\displaystyle 51.5 (D) 52\displaystyle 52

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 51\). \[\sum x_i = 100\times50 = 5000 \] \[\text{New sum} = 5000-50+150 = 5100 \] \[\text{New mean} = \frac{5100}{100} = 51 \]

  9. Exercise 19

    There are 50\displaystyle 50 numbers. Each number is subtracted from 53\displaystyle 53 and the mean of the numbers so obtained is found to be -3.5. The mean of the given numbers is : (A) 46.5\displaystyle 46.5 (B) 49.5\displaystyle 49.5 (C) 53.5\displaystyle 53.5 (D) 56.5\displaystyle 56.5

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 56.5\). \[y_i = 53-x_i \quad\Rightarrow\quad \bar{y} = 53-\bar{x} \] \[-3.5 = 53-\bar{x} \] \[\bar{x} = 53+3.5 = 56.5 \]

  10. Exercise 20

    The mean of 25\displaystyle 25 observations is 36. Out of these observations if the mean of first 13\displaystyle 13 observations is 32\displaystyle 32 and that of the last 13\displaystyle 13 observations is 40\displaystyle 40 , the 13th \displaystyle 13^{\text {th }} observation is: (A) 23\displaystyle 23 (B) 36\displaystyle 36 (C) 38\displaystyle 38 (D) 40\displaystyle 40

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 36\). \[\sum_{1}^{25} x_i = 25\times36 = 900 \] \[\sum_{1}^{13} x_i = 13\times32 = 416, \qquad \sum_{13}^{25} x_i = 13\times40 = 520 \] \[x_{13} = 416+520-900 = 36 \]