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NCERT Exemplar · Class 11 Mathematics Statistics

46 questions · 46 still being checked

EXERCISE 15.3 41–46 (part 5 of 5)

  1. Fill in the blanks in Exercises from $\displaystyle 40$ to 46.

    Exercise 41

    If xˉ\displaystyle \bar{x} is the mean of n\displaystyle n values of x\displaystyle x, then ∑i=1n(xi−xˉ)\displaystyle \sum_{i=1}^n\left(x_i-\bar{x}\right) is always equal to ____\displaystyle \_\_\_\_. If a\displaystyle a has any value other than xˉ\displaystyle \bar{x}, then ∑i=1n(xi−xˉ)2\displaystyle \sum_{i=1}^n\left(x_i-\bar{x}\right)^2 is ____\displaystyle \_\_\_\_ than ∑(xi−a)2\displaystyle \sum\left(x_i-a\right)^2

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    $\displaystyle 0$, less
    \(\displaystyle 0\); less\[\sum_{i=1}^n (x_i-\bar{x}) = \sum x_i - n\bar{x} = n\bar{x}-n\bar{x} = 0 \] \[x_i-a = (x_i-\bar{x}) + (\bar{x}-a) \] \[\sum (x_i-a)^2 = \sum (x_i-\bar{x})^2 + 2(\bar{x}-a)\sum (x_i-\bar{x}) + n(\bar{x}-a)^2 \] \[\sum (x_i-a)^2 = \sum (x_i-\bar{x})^2 + n(\bar{x}-a)^2 \] \[n(\bar{x}-a)^2 > 0 \quad \text{(as } a\neq \bar{x}\text{)} \] \[\sum (x_i-\bar{x})^2 < \sum (x_i-a)^2 \]Answer: \(\displaystyle 0\); less

  2. Exercise 42

    If the variance of a data is 121\displaystyle 121, then the standard deviation of the data is ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    $\displaystyle 11$
    \(\displaystyle 11\)\[\sigma = \sqrt{\text{variance}} \] \[\sigma = \sqrt{121} = 11 \]Answer: \(\displaystyle 11\)

  3. Exercise 43

    The standard deviation of a data is ____\displaystyle \_\_\_\_ of any change in orgin, but is ____\displaystyle \_\_\_\_ on the change of scale.

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    NCERT’s answer
    Independent
    independent; dependentChange origin and scale together: \(\displaystyle y_i = px_i + q\).\[\bar{y} = p\bar{x}+q \] \[y_i-\bar{y} = p(x_i-\bar{x}) \] \[\sigma_y^2 = \frac{1}{n}\sum p^2(x_i-\bar{x})^2 = p^2\sigma_x^2 \] \[\sigma_y = |p|\,\sigma_x \]\(\displaystyle q\) (origin) does not appear, so \(\displaystyle \sigma\) is unchanged; \(\displaystyle p\) (scale) multiplies it by \(\displaystyle |p|\).Answer: independent; dependent

  4. Exercise 44

    The sum of the squares of the deviations of the values of the variable is ____\displaystyle \_\_\_\_ when taken about their arithmetic mean.

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    NCERT’s answer
    Minimum
    least\[\sum (x_i-a)^2 = \sum (x_i-\bar{x})^2 + n(\bar{x}-a)^2 \] \[n(\bar{x}-a)^2 \ge 0 \] \[\sum (x_i-a)^2 \ge \sum (x_i-\bar{x})^2 \]Equality holds only for \(\displaystyle a=\bar{x}\).Answer: least (minimum)

  5. Exercise 45

    The mean deviation of the data is ____\displaystyle \_\_\_\_ when measured from the median.

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    NCERT’s answer
    Least
    leastArrange \(\displaystyle x_1\le x_2\le\cdots\le x_n\). For \(\displaystyle i\le n/2\):\[|x_i-a| + |x_{n+1-i}-a| \ge x_{n+1-i}-x_i \quad \text{(triangle inequality)} \] \[\sum_{i=1}^n |x_i-a| \ge \sum_{i\le n/2}\left(x_{n+1-i}-x_i\right) \]The right side does not involve \(\displaystyle a\). Equality holds when \(\displaystyle x_i\le a\le x_{n+1-i}\) for every \(\displaystyle i\le n/2\) (and \(\displaystyle a=x_{(n+1)/2}\) if \(\displaystyle n\) is odd). The median satisfies this.Answer: least (minimum)

  6. Exercise 46

    The standard deviation is ____\displaystyle \_\_\_\_ to the mean deviation taken from the arithmetic mean.

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    NCERT’s answer
    greater than or equal
    greater than or equal toPut \(\displaystyle d_i=|x_i-\bar{x}|\). The variance of the numbers \(\displaystyle d_i\) is non-negative:\[\frac{1}{n}\sum d_i^2 - \left(\frac{1}{n}\sum d_i\right)^2 \ge 0 \] \[\sigma^2 = \frac{1}{n}\sum (x_i-\bar{x})^2 = \frac{1}{n}\sum d_i^2 \] \[\sigma^2 \ge \left(\frac{1}{n}\sum |x_i-\bar{x}|\right)^2 = (\text{M.D.})^2 \] \[\sigma \ge \text{M.D.} \]Equality holds only when all \(\displaystyle d_i\) are equal.Answer: greater than or equal to