Let the observation $x _ { i } ( 1 \leq i \leq 10 )$ satisfy the equations $\sum _ { i = 1 } ^ { 10 } \left( x _ { i } - 5 \right) = 10 , \sum _ { i = 1 } ^ { 10 } \left( x _ { i } - 5 \right) ^ { 2 } = 40$. If $\mu$ and $\lambda$ are the mean and the variance of the observations, $x _ { 1 } - 3 , x _ { 2 } - 3 , \ldots , x _ { 10 } - 3$, then the ordered pair $( \mu , \lambda )$ is equal to:
(1) $( 3,3 )$
(2) $( 6,3 )$
(3) $( 6,6 )$
(4) $( 3,6 )$

For the frequency distribution: Variate $( x ) : x _ { 1 } , x _ { 2 } , x _ { 3 } , \ldots , x _ { 15 }$ Frequency $( f ) : f _ { 1 } , f _ { 2 } , f _ { 3 } , \ldots , f _ { 15 }$ where $0 < x _ { 1 } < x _ { 2 } < x _ { 3 } < \ldots < x _ { 15 } = 10$ and $\sum _ { i = 1 } ^ { 15 } f _ { i } > 0$, the standard deviation cannot be
(1) 4
(2) 1
(3) 6
(4) 2

The mean and variance of 8 observations are 10 and 13.5 respectively. If 6 of these observations are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is:
(1) 7
(2) 3
(3) 5
(4) 9

The mean and variance of 7 observations are 8 and 16, respectively. If five observations are $2, 4, 10, 12, 14$ then the absolute difference of the remaining two observations is :
(1) 1
(2) 4
(3) 2
(4) 3

If $\sum _ { i = 1 } ^ { n } \left( x _ { i } - a \right) = n$ and $\sum _ { i = 1 } ^ { n } \left( x _ { i } - a \right) ^ { 2 } = n a , ( n , a > 1 )$, then the standard deviation of $n$ observations $x _ { 1 } , x _ { 2 } , \ldots , x _ { n }$ is
(1) $a - 1$
(2) $n \sqrt { ( a - 1 ) }$
(3) $\sqrt { n ( a - 1 ) }$
(4) $\sqrt { ( a - 1 ) }$

If the mean and the standard deviation of the data $3, 5, 7, a, b$ are 5 and 2 respectively, then $a$ and $b$ are the roots of the equation:
(1) $x^2 - 10x + 18 = 0$
(2) $2x^2 - 20x + 19 = 0$
(3) $x^2 - 10x + 19 = 0$
(4) $x^2 - 20x + 18 = 0$

If the variance of the following frequency distribution:

Class:	$10 - 20$	$20 - 30$	$30 - 40$
Frequency:	2	$x$	2

is 50, then $x$ is equal to $\_\_\_\_$

If the variance of the first $n$ natural numbers is 10 and the variance of the first $m$ even natural numbers is 16, then the value of $m + n$ is equal to

The mean of 6 distinct observations is 6.5 and their variance is 10.25 . If 4 out of 6 observations are $2,4,5$ and 7, then the remaining two observations are:
(1) 10,11
(2) 3,18
(3) 8,13
(4) 1,20

If the mean and variance of six observations $7,10,11,15 , a , b$ are 10 and $\frac { 20 } { 3 }$, respectively, then the value of $| a - b |$ is equal to:
(1) 9
(2) 11
(3) 7
(4) 1

Let in a series of $2 n$ observations, half of them are equal to $a$ and remaining half are equal to $- a$. Also by adding a constant $b$ in each of these observations, the mean and standard deviation of new set become 5 and 20 , respectively. Then the value of $a ^ { 2 } + b ^ { 2 }$ is equal to:
(1) 425
(2) 650
(3) 250
(4) 925

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