Consider three observations $a , b$ and $c$ such that $b = a + c$. If the standard deviation of $a + 2 , b + 2 , c + 2$ is $d$, then which of the following is true?
(1) $b ^ { 2 } = 3 \left( a ^ { 2 } + c ^ { 2 } \right) + 9 d ^ { 2 }$
(2) $b ^ { 2 } = a ^ { 2 } + c ^ { 2 } + 3 d ^ { 2 }$
(3) $b ^ { 2 } = 3 \left( a ^ { 2 } + c ^ { 2 } + d ^ { 2 } \right)$
(4) $b ^ { 2 } = 3 \left( a ^ { 2 } + c ^ { 2 } \right) - 9 d ^ { 2 }$

The first of the two samples in a group has 100 items with mean 15 and standard deviation 3. If the whole group has 250 items with mean 15.6 and standard deviation $\sqrt { 13.44 }$, then the standard deviation of the second sample is:
(1) 8
(2) 6
(3) 4
(4) 5

Consider a set of $3 n$ numbers having variance 4 . In this set, the mean of first $2 n$ numbers is 6 and the mean of the remaining $n$ numbers is 3 . A new set is constructed by adding 1 into each of the first $2 n$ numbers, and subtracting 1 from each of the remaining $n$ numbers. If the variance of the new set is $k$, then $9 k$ is equal to $\_\_\_\_$ .

The mean of 10 numbers $7 \times 8, 10 \times 10, 13 \times 12, 16 \times 14 , \ldots$ is

An online exam is attempted by 50 candidates out of which 20 are boys. The average marks obtained by boys is 12 with a variance 2 . The variance of marks obtained by 30 girls is also 2 . The average marks of all 50 candidates is 15 . If $\mu$ is the average marks of girls and $\sigma ^ { 2 }$ is the variance of marks of 50 candidates, then $\mu + \sigma ^ { 2 }$ is equal to

Let the mean of 50 observations is 15 and the standard deviation is 2. However, one observation was wrongly recorded. The sum of the correct and incorrect observations is 70. If the mean of the correct set of observations is 16, then the variance of the correct set is equal to
(1) 10
(2) 36
(3) 43
(4) 60

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 observation are $\frac { 7 } { 2 }$ and $a$ respectively, then ( $4 a + x _ { 5 }$ ) is equal to
(1) 13
(2) 15
(3) 17
(4) 18

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 $25M$ is equal to
(1) 60
(2) 55
(3) 50
(4) 75

The mean and variance of the data $4,5,6,6,7,8 , x , y$ where $x < y$ are 6 and $\frac { 9 } { 4 }$ respectively. Then $x ^ { 4 } + y ^ { 2 }$ is equal to
(1) 320
(2) 420
(3) 162
(4) 674

If the mean deviation about median for the number $3, 5, 7, 2k, 12, 16, 21, 24$ arranged in the ascending order, is 6 then the median is
(1) 11.5
(2) 10.5
(3) 12
(4) 11

The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then, the correct standard deviation 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
(1) 1072
(2) 1792
(3) 1216
(4) 1456

The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is
(1) 11
(2) 13
(3) 12
(4) 14

Let $\mu$ be the mean and $\sigma$ be the standard deviation of the distribution

$X _ { i }$	0	1	2	3	4	5
$f _ { i }$	$k + 2$	$2k$	$k ^ { 2 } - 1$	$k ^ { 2 } - 1$	$k ^ { 2 } + 1$	$k - 3$

where $\Sigma f _ { i } = 62$. If $\lfloor x \rfloor$ denotes the greatest integer $\leq x$, then $\lfloor \mu ^ { 2 } + \sigma ^ { 2 } \rfloor$ is equal to
(1) 9
(2) 8
(3) 7
(4) 6

The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then the correct mean and variance are
(1) 14 and 13.5
(2) 14 and 12.5
(3) 15 and 14.5
(4) 14 and 11.5

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
(1) 500
(2) 450
(3) 650
(4) 900

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:
(1) 4.04
(2) 4.08
(3) 3.96
(4) 3.92

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
(1) $\phi$
(2) $\{99\}$
(3) $\mathbb{N}$
(4) $\{9\}$

Let the positive numbers $a _ { 1 } , a _ { 2 } , a _ { 3 } , a _ { 4 }$ and $a _ { 5 }$ be in a G.P. Let their mean and variance be $\frac { 31 } { 10 }$ and $\frac { m } { n }$ respectively, where $m$ and $n$ are co-prime. If the mean of their reciprocals is $\frac { 31 } { 10 }$ and $a _ { 3 } + a _ { 4 } + a _ { 5 } = 14$, then $m + n$ is equal to $\_\_\_\_$ .

If the mean and variance of the frequency distribution

$x_{i}$	2	4	6	8	10	12	14	16
$f_{i}$	4	4	$\alpha$	15	8	$\beta$	4	5

are 9 and 15.08 respectively, then the value of $\alpha^{2} + \beta^{2} - \alpha\beta$ is $\_\_\_\_$.

Let the mean and variance of 12 observations be $\frac { 9 } { 2 }$ and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. Find the correct mean and variance.

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

Let the mean and variance of 8 numbers $x , y , 10 , 12 , 6 , 12 , 4 , 8$ be 9 and 9.25 respectively. If $x > y$, then $3 x - 2 y$ is equal to $\_\_\_\_$

If the variance of the frequency distribution

$x_i$	2	3	4	5	6	7	8
Frequency $f_i$	3	6	16	$\alpha$	9	5	6

is 3, then $\alpha$ is equal to $\underline{\hspace{1cm}}$.

If the mean of the frequency distribution

Class :	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequency :	2	3	$x$	5

is 28, then its variance is $\_\_\_\_$.