One set containing five numbers has mean $8$ and variance $18$,and the second set containing $3$ numbers has mean $8$ and variance $24$. Then the variance of the combined set of numbers is

  • A
    $42$
  • B
    $20.25$
  • C
    $18$
  • D
    None of these

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Similar Questions

Let the mean of the data be $5$.
$X$ $1$ $3$ $5$ $7$ $9$
$f$ $4$ $24$ $28$ $\alpha$ $8$
If $m$ and $\sigma^2$ are respectively the mean deviation about the mean and the variance of the data,then $\frac{3 \alpha}{m+\sigma^2}$ is equal to $..........$.

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 $(4) \ 145$
$(5) \ 55$

Let the mean and variance of $8$ numbers $x, y, 10, 12, 6, 12, 4, 8$ be $9$ and $9.25$ respectively. If $x > y$,then $3x - 2y$ is equal to $...........$.

Let $X = \{11, 12, 13, \ldots, 40, 41\}$ and $Y = \{61, 62, 63, \ldots, 90, 91\}$ be two sets of observations. If $\bar{x}$ and $\bar{y}$ are their respective means and $\sigma^2$ is the variance of all the observations in $X \cup Y$,then $|\bar{x} + \bar{y} - \sigma^2|$ is equal to $.................$.

For two data sets,each of size $5$,the variances are given to be $4$ and $5$,and the corresponding means are given to be $2$ and $4$ respectively. The variance of the combined data set is

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