Variance and standard deviation

Consider a data consisting of 10 observations $x_1,x_2,\dots,x_{10}$, whose mean is $5$ and variance is $7$. If the mean and the variance of the first $8$ observations $x_1,x_2,\dots,x_8$ are $4$ and $3.5$, respectively, and $x_9<x_{10}$, then the value of $3x_9+2x_{10}$ is ________.

The mean and variance of $n$ observations are 8 and 16, respectively. If the sum of the first $(n-1)$ observations is 48 and the sum of squares of the first $(n-1)$ observations is 496, then the value of $n$ is :

Suppose that the mean and median of the non-negative numbers 21, 8, 17, $a$, 51, 103, $b$, 13, 67, $(a>b)$, are 40 and 21, respectively. If the mean deviation about the median is 26, then $2a$ is equal to:

For 10 observations $x_1,x_2,\dots,x_{10}$, if $\displaystyle\sum_{i=1}^{10}(x_i+2)^2=180$ and $\displaystyle\sum_{i=1}^{10}(x_i-1)^2=90$, then their standard deviation is:

The mean deviation about the mean for the data

$x_i$	5	7	9	10	12	15
$f_i$	8	6	2	2	2	6

is equal to:

A data consists of 20 observations $x_1,x_2,\dots,x_{20}$. If $\displaystyle\sum_{i=1}^{20}(x_i+5)^2=2500$ and $\displaystyle\sum_{i=1}^{20}(x_i-5)^2=100$, then the ratio of mean to standard deviation of this data is:

Let the mean and the variance of seven observations $2,4,\alpha,8,\beta,12,14$, $\alpha<\beta$, be 8 and 16 respectively. Then the quadratic equation whose roots are $3\alpha+2$ and $2\beta+1$ is :

A set of four observations has mean 1 and variance 13. Another set of six observations has mean 2 and variance 1. Then, the variance of all these 10 observations is equal to :

Consider the following frequency distribution:

Value	4	5	8	9	6	12	11
Frequency	5	$f_1$	$f_2$	2	1	1	3

Suppose that the sum of the frequencies is 19 and the median of this frequency distribution is 6.

For the given frequency distribution, let $\alpha$ denote the mean deviation about the mean, $\beta$ denote the mean deviation about the median, and $\sigma^2$ denote the variance.

Match each entry in List-I to the correct entry in List-II and choose the correct option.

Let $X$ be a random variable, and let $P(X = x)$ denote the probability that $X$ takes the value $x$. Suppose that the points $(x, P(X = x))$, $x = 0, 1, 2, 3, 4$, lie on a fixed straight line in the $xy$-plane, and $P(X = x) = 0$ for all $x \in \mathbb{R} - \{0, 1, 2, 3, 4\}$. If the mean of $X$ is $\dfrac{5}{2}$, and the variance of $X$ is $\alpha$, then the value of $24\alpha$ is _____.

Consider the given data with frequency distribution

$x_i$	3	8	11	10	5	4
$f_i$	5	2	3	2	4	4

Match each entry in List-I to the correct entries in List-II.

The correct option is:

If the mean deviation about the mean of the numbers 1, 2, 3, …, $n$, where $n$ is odd, is $\dfrac{5(n+1)}n$, then $n$ 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 :

The mean and variance of the data 4, 5, 6, 6, 7, 8, $x$, $y$, where $x<y$, are 6 and $\dfrac94$ respectively. Then $x^4+y^2$ is equal to

The mean and standard deviation of 15 observations are found to be 8 and 3 respectively. On rechecking it was found that, in the observations, 20 was misread as 5. Then, the correct variance is equal to ______.

Suppose a class has 7 students. The average marks of these students in the mathematics examination is 62, and their variance is 20. A student fails in the examination if he/she gets less than 50 marks, then in worst case, the number of students can fail is ______.

Let the mean and the variance of 5 observations $x_1,x_2,x_3,x_4,x_5$ be $\dfrac{24}5$ and $\dfrac{194}{25}$ respectively. If the mean and variance of the first 4 observation are $\dfrac72$ and $a$ respectively, then $(4a+x_5)$ is equal to :

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

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