The number of values of $a \in N$ such that the variance of $3,7,12 a, 43-a$ is a natural number is (Mean $=13$)
$0$
$2$
$5$
infinite
Let the mean and the variance of $5$ observations $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be $\frac{24}{5}$ and $\frac{194}{25}$ respectively. If the mean and variance of the first $4$ observation are $\frac{7}{2}$ and $a$ respectively, then $\left(4 a+x_{5}\right)$ is equal to
Find the mean, variance and standard deviation using short-cut method
Height in cms | $70-75$ | $75-80$ | $80-85$ | $85-90$ | $90-95$ | $95-100$ | $100-105$ | $105-110$ | $110-115$ |
No. of children | $3$ | $4$ | $7$ | $7$ | $15$ | $9$ | $6$ | $6$ | $3$ |
Mean and variance of a set of $6$ terms is $11$ and $24$ respectively and the mean and variance of another set of $3$ terms is $14$ and $36$ respectively. Then variance of all $9$ terms is equal to
The mean and the standard deviation $(s.d.)$ of five observations are $9$ and $0,$ respectively. If one of the observations is changed such that the mean of the new set of five observations becomes $10,$ then their $s.d.$ is?
The mean and standard deviation of the marks of $10$ students were found to be $50$ and $12$ respectively. Later, it was observed that two marks $20$ and $25$ were wrongly read as $45$ and $50$ respectively. Then the correct variance is $............$.