The mean and standard deviation of marks obtained by $50$ students of a class in three subjects, Mathematics, Physics and Chemistry are given below:
Subject | Mathematics | Physics | Chemistty |
Mean | $42$ | $32$ | $40.9$ |
Standard deviation | $12$ | $15$ | $20$ |
Which of the three subjects shows the highest variability in marks and which shows the lowest?
Standard deviation of Mathematics $=12$
Standard deviation of Physics $=15$
Standard deviation of Chemistry $=20$
The coefficient of variation $( C.V. )$ is given by $\frac{\text { Standard deviation }}{\text { Mean }} \times 100$
$C.V.$ (in Mathematics) $=\frac{12}{42} \times 100=28.57$
$C.V.$ (in Physics) $=\frac{15}{32} \times 100=46.87$
$C.V.$ (in Chemistry) $=\frac{20}{40.9} \times 100=48.89$
The subject with greater $C.V.$ is more variable than others.
Therefore, the highest variability in marks is in Chemistry and the lowest variability in marks is in Mathematics.
Let $x_1, x_2, x_3, x_4, .......... , x_n$ be $n$ observations and let $\bar x$ be their arithmetic mean and $\sigma ^2$ be their variance.
Statement $-1$ : Variance of observations $2x_1, 2x_2, 2x_3, ......, 2x_n$ is $4\sigma ^2$ .
Statement $-2$ : Arithmetic mean of $2x _1, 2x_2, 2x_3, ......, 2x_n$ is $4\bar x$ .
The variance of $20$ observations is $5 .$ If each observation is multiplied by $2,$ find the new variance of the resulting observations.
The mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours. Find the overall standard deviation.
Find the mean and variance for the following frequency distribution.
Classes | $0-30$ | $30-60$ | $60-90$ | $90-120$ | $120-150$ | $50-180$ | $180-210$ |
$f_i$ | $2$ | $3$ | $5$ | $10$ | $3$ | $5$ | $2$ |
Let $x_1, x_2,........,x_n$ be $n$ observations such that $\sum {{x_i}^2 = 300} $ and $\sum {{x_i} = 60} $ on value of $n$ among the following is