The mean and variance of the marks obtained by the students in a test are $10$ and $4$ respectively. Later, the marks of one of the students is increased from $8$ to $12$ . If the new mean of the marks is $10.2.$ then their new variance is equal to :
$4.04$
$4.08$
$3.96$
$3.92$
Calculate the mean, variance and standard deviation for the following distribution:
Class | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ | $80-90$ | $90-100$ |
$f_i$ | $3$ | $7$ | $12$ | $15$ | $8$ | $3$ | $2$ |
From a lot of $12$ items containing $3$ defectives, a sample of $5$ items is drawn at random. Let the random variable $\mathrm{X}$ denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of $X$ is $\frac{m}{n}$, where $\operatorname{gcd}(m, n)=1$, then $n-m$ is equal to..........
Let the six numbers $a_1, a_2, a_3, a_4, a_5, a_6$ be in $A.P.$ and $a_1+a_3=10$. If the mean of these six numbers is $\frac{19}{2}$ and their variance is $\sigma^2$, then $8 \sigma^2$ is equal to
The mean and $S.D.$ of the marks of $200$ candidates were found to be $40$ and $15$ respectively. Later, it was discovered that a score of $40$ was wrongly read as $50$. The correct mean and $S.D.$ respectively are...