Find the mean and variance for the first $10$ multiples of $3$
The first $10$ multiples of $3$ are
$3,6,9,12,15,18,21,24,27,30$
Here, number of observations, $n=10$
Mean, $\bar x = \frac{{\sum\limits_{i = 1}^{10} {{x_i}} }}{{10}} = \frac{{165}}{{10}} = 16.5$
The following table is obtained.
${x_i}$ | $\left( {{x_i} - \bar x} \right)$ | ${\left( {{x_i} - \bar x} \right)^2}$ |
$3$ | $-13.5$ | $182.25$ |
$6$ | $-10.5$ | $110.25$ |
$9$ | $-7.5$ | $56.25$ |
$12$ | $-4.5$ | $20.25$ |
$15$ | $-1.5$ | $2.25$ |
$18$ | $1.5$ | $2.25$ |
$21$ | $4.5$ | $20.25$ |
$24$ | $7.5$ | $56.25$ |
$27$ | $10.5$ | $110.25$ |
$30$ | $13.5$ | $182.25$ |
Variance $\left( {{\sigma ^2}} \right) = \frac{1}{n}\sum\limits_{i = 1}^{10} {{{\left( {{x_1} - \bar x} \right)}^2} = } \frac{1}{{10}} \times 742.5 = 74.25$
$742.5$
If the variance of the first $n$ natural numbers is $10$ and the variance of the first m even natural numbers is $16$, then $m + n$ is equal to
The $S.D$. of the first $n$ natural numbers is
Calculate mean, variance and standard deviation for the following distribution.
Classes | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ | $80-90$ | $90-100$ |
${f_i}$ | $3$ | $7$ | $12$ | $15$ | $8$ | $3$ | $2$ |
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 $=13$)
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 $............$.