The mean and standard deviation of $100$ observations were calculated as $40$ and $5.1$ respectively. Later on,it was found that one of the observations was taken as $50$ in place of $40$. If the wrong entry is replaced by the correct one,then the sum of the squares of all the observations is:
- A$162701$
- B$163501$
- C$162601$
- D$161701$
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The mean and standard deviation of $20$ observations are found to be $10$ and $2$ respectively. On rechecking,it was found that an observation $8$ was incorrect. Calculate the correct mean and standard deviation if the wrong item is omitted.
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View Solution$x_1, x_2, \ldots, x_n$ are $n$ observations with mean $\bar{x}$ and standard deviation $\sigma$. Match the items of List-$I$ with those of List-$II$:
List-$I$ List-$II$ $(a) \sum_{i=1}^n(x_i-\bar{x})$ $(i) \text{ Median}$ $(b) \text{ Variance } (\sigma^2)$ $(ii) \text{ Coefficient of variation}$ $(c) \text{ Mean deviation}$ $(iii) \text{ Zero}$ $(d) \text{ Measure used to find the homogeneity of given two series}$ $(iv) \text{ Mean of the absolute deviations from any measure of central tendency}$ $(v) \text{ Mean of the squares of the deviations from mean}$
| List-$I$ | List-$II$ |
|---|---|
| $(a) \sum_{i=1}^n(x_i-\bar{x})$ | $(i) \text{ Median}$ |
| $(b) \text{ Variance } (\sigma^2)$ | $(ii) \text{ Coefficient of variation}$ |
| $(c) \text{ Mean deviation}$ | $(iii) \text{ Zero}$ |
| $(d) \text{ Measure used to find the homogeneity of given two series}$ | $(iv) \text{ Mean of the absolute deviations from any measure of central tendency}$ |
| $(v) \text{ Mean of the squares of the deviations from mean}$ |
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