Statistics

Let the mean and the variance of 5 observations x₁, x₂, x₃, x₄, x₅ be ${24 \over 5}$ and ${194 \over 25}$ respectively. If the mean and variance of the first 4 observation are ${7 \over 2}$ and a respectively, then (4a + x₅) is equal to:

The number of values of a $\in$ N such that the variance of 3, 7, 12, a, 43 $-$ a is a natural number is :

The mean and variance of the data 4, 5, 6, 6, 7, 8, x, y, where x < y, are 6 and ${9 \over 4}$ respectively. Then ${x^4} + {y^2}$ is equal to :

The mean of the numbers a, b, 8, 5, 10 is 6 and their variance is 6.8. If M is the mean deviation of the numbers about the mean, then 25 M is equal to :

The mean and standard deviation of 50 observations are 15 and 2 respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70. If the correct mean is 16, then the correct variance is equal to :

If the mean deviation about median for the numbers 3, 5, 7, 2k, 12, 16, 21, 24, arranged in the ascending order, is 6 then the median is :

Let $9=x_{1} < x_{2} < \ldots < x_{7}$ be in an A.P. with common difference d. If the standard deviation of $x_{1}, x_{2}..., x_{7}$ is 4 and the mean is $\bar{x}$, then $\bar{x}+x_{6}$ is equal to :

Let the mean and standard deviation of marks of class A of 100 students be respectively 40 and $\alpha(>$ 0 ), and the mean and standard deviation of marks of class $B$ of $n$ students be respectively 55 and 30 $-\alpha$. If the mean and variance of the marks of the combined class of $100+\mathrm{n}$ studants are respectively 50 and 350 , then the sum of variances of classes $A$ and $B$ is :

The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is :

Let $S$ be the set of all values of $a_1$ for which the mean deviation about the mean of 100 consecutive positive integers $a_1, a_2, a_3, \ldots ., a_{100}$ is 25 . Then $S$ is :