MCA Entrance Previous Year Question Bank

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

Let the mean and the variance of $5$ observations $x_1,x_2,x_3,x_4,x_5$ be $\frac{24}{5}$ and $\frac{194}{25}$ respectively. If the mean and variance of the first $4$ observations are $\frac{7}{2}$ and $a$ respectively, then $(4a+x_5)$ is equal to:

Let 9 = x_1 < x_2 < \dots < x_7 be in an A.P. with common difference $d$. If the standard deviation of $x_1,x_2,\dots ,x_7$ is $4$ and mean is $\bar{x}$, then $\bar{x}+x_6$ is equal to:

In a binomial distribution $B(n,p)$, the sum and the product of the mean and the variance are 5 and 6 respectively, then $6(n+p-q)$ 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 $100+n$ students are respectively $50$ and $350$, then the sum of variances of classes A and B is :

If the variance of the frequency distribution

is 3, then a is equal to

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

The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted and $a$ and $b$ are respectively mean and variance of remaining 6 observations, then $a+3b-5$ is equal to

Let $X=\{11,12,13,\dots ,40,41\}$ and $Y=\{61,62,63,\dots ,90,91\}$ be the two sets of observations. If $\bar{x}$ and $\bar{y}$ are their respective means and ${\sigma }^2$ is the variance of all the observations in $X\cup Y$, then $|\bar{x}+\bar{y}-{\sigma }^2|$ is equal to:

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: