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 | $3$ | $4$ | $7$ | $7$ | $15$ | $9$ | $6$ | $6$ | $3$ |
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
Suppose a class has $7$ students. The average marks of these students in the mathematics examination is $62$, and their variance is $20$ . A student fails in the examination if $he/she$ gets less than $50$ marks, then in worst case, the number of students can fail is
In any discrete series (when all values are not same) the relationship between $M.D.$ about mean and $S.D.$ is
The sum of squares of deviations for $10$ observations taken from mean $50$ is $250$. The co-efficient of variation is.....$\%$