Consider the following frequency distribution:
Value $4$ $5$ $8$ $9$ $6$ $12$ $11$
Frequency $5$ $f_1$ $f_2$ $2$ $1$ $1$ $3$

Suppose that the sum of the frequencies is $19$ and the median of this frequency distribution is $6$. For the given frequency distribution,let $\alpha$ denote the mean deviation about the mean,$\beta$ denote the mean deviation about the median,and $\sigma^2$ denote the variance. Match each entry in List-$I$ to the correct entry in List-$II$ and choose the correct option.
List-$I$ List-$II$
$(P) \ 7f_1+9f_2$ is equal to $(1) \ 146$
$(Q) \ 19\alpha$ is equal to $(2) \ 47$
$(R) \ 19\beta$ is equal to $(3) \ 48$
$(S) \ 19\sigma^2$ is equal to $(4) \ 145$
$(5) \ 55$

  • A
    $(P) \rightarrow (5), (Q) \rightarrow (3), (R) \rightarrow (2), (S) \rightarrow (4)$
  • B
    $(P) \rightarrow (5), (Q) \rightarrow (2), (R) \rightarrow (3), (S) \rightarrow (1)$
  • C
    $(P) \rightarrow (5), (Q) \rightarrow (3), (R) \rightarrow (2), (S) \rightarrow (1)$
  • D
    $(P) \rightarrow (3), (Q) \rightarrow (2), (R) \rightarrow (5), (S) \rightarrow (4)$

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The mean and standard deviation of $15$ observations were found to be $12$ and $3$ respectively. On rechecking,it was found that an observation was read as $10$ in place of $12$. If $\mu$ and $\sigma^2$ denote the mean and variance of the correct observations respectively,then $15(\mu+\mu^2+\sigma^2)$ is equal to $...................$

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For a group of $100$ observations,the arithmetic mean and standard deviation are $8$ and $\sqrt{10.5}$ respectively. The mean and standard deviation of $50$ items selected from these $100$ observations are $10$ and $2$ respectively. Then the standard deviation of the remaining $50$ observations is

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An analysis of monthly wages paid to the workers of two jute mills $A$ and $B$ gives the following data:
Metric Mill-$A$ Mill-$B$
No. of workers $500$ $600$
Average daily wage (in rupees) $186$ $175$
Variance of distribution of wages $81$ $100$

Then:

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