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.
The mean and standard deviation of $20$ observations are found to be $10$ and $2$ respectively. On rechecking, it was found that an observation $8$ was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
If wrong item is omitted.
Let $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{N}$ and $\mathrm{a}<\mathrm{b}<\mathrm{c}$. Let the mean, the mean deviation about the mean and the variance of the $5$ observations $9$,$25$, $a$, $b$, $c$ be $18$,$4$ and $\frac{136}{5}$, respectively. Then $2 \mathrm{a}+\mathrm{b}-\mathrm{c}$ is equal to ..............
Let $\mathrm{X}$ be a random variable with distribution.
$\mathrm{x}$ | $-2$ | $-1$ | $3$ | $4$ | $6$ |
$\mathrm{P}(\mathrm{X}=\mathrm{x})$ | $\frac{1}{5}$ | $\mathrm{a}$ | $\frac{1}{3}$ | $\frac{1}{5}$ | $\mathrm{~b}$ |
If the mean of $X$ is $2.3$ and variance of $X$ is $\sigma^{2}$, then $100 \sigma^{2}$ is equal to :
Mean of $5$ observations is $7.$ If four of these observations are $6, 7, 8, 10$ and one is missing then the variance of all the five observations 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 $............$.