The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If the mean age of the teachers in this school now is 39 years, then the age (in years) of the newly appointed teacher is:
(1) 25
(2) 30
(3) 35
(4) 40

If $\sum _ { i = 1 } ^ { 9 } ( x _ { i } - 5 ) = 9$ and $\sum _ { i = 1 } ^ { 9 } ( x _ { i } - 5 ) ^ { 2 } = 45$, then the standard deviation of the 9 items $x _ { 1 } , x _ { 2 } , \ldots , x _ { 9 }$ is:
(1) 9
(2) 4
(3) 2
(4) 3

The mean of a set of 30 observation is 75. If each observations is multiplied by non-zero number $\lambda$ and then each of them is decreased by 25, their mean remains the same. Then, $\lambda$ is equal to :
(1) $\frac { 4 } { 3 }$
(2) $\frac { 1 } { 3 }$
(3) $\frac { 10 } { 3 }$
(4) $\frac { 2 } { 3 }$

The mean and the standard deviation (S. D.) of five observations are 9 and 0 , respectively. If one of the observation is increased such that the mean of the new set of five observations becomes 10 , then their S. D. is
(1) 0
(2) 2
(3) 4
(4) 1

The mean of a set of 30 observations is 75 . If each other observation is multiplied by a nonzero number $\lambda$ and then each of them is decreased by 25 , their mean remains the same. The $\lambda$ is equal to
(1) $\frac { 10 } { 3 }$
(2) $\frac { 4 } { 3 }$
(3) $\frac { 1 } { 3 }$
(4) $\frac { 2 } { 3 }$

If $\sum _ { i = 1 } ^ { 9 } \left( x _ { i } - 5 \right) = 9$ and $\sum _ { i = 1 } ^ { 9 } \left( x _ { i } - 5 \right) ^ { 2 } = 45$, then the standard deviation of the 9 items $x _ { 1 } , x _ { 2 } , \ldots , x _ { 9 }$ is
(1) 3
(2) 9
(3) 4
(4) 2

If the standard deviation of the numbers $- 1,0,1 , k$ is $\sqrt { 5 }$ where $k > 0$, then $k$ is equal to
(1) $\sqrt { 6 }$
(2) $4 \sqrt { \frac { 5 } { 3 } }$
(3) $2 \sqrt { \frac { 10 } { 3 } }$
(4) $2 \sqrt { 6 }$

The mean and the variance of five observations are 4 and 5.20 , respectively. If three of the observations are 3, 4 and 4 ; then the absolute value of the difference of the other two observations, is :
(1) 3
(2) 5
(3) 7
(4) 1

A student scores the following marks in five tests: $45,54,41,57,43$. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is:
(1) $\frac { 10 } { 3 }$
(2) $\frac { 100 } { 3 }$
(3) $\frac { 10 } { \sqrt { 3 } }$
(4) $\frac { 100 } { \sqrt { 3 } }$

The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is
(1) $10 : 3$
(2) $4 : 9$
(3) $6 : 7$
(4) $5 : 8$

The mean and the median of the following ten numbers in increasing order $10,22,26,29,34 , x , 42,67,70 , y$ are 42 and 35 respectively, then $\frac { y } { x }$ is equal to:
(1) $\frac { 9 } { 4 }$
(2) $\frac { 7 } { 3 }$
(3) $\frac { 7 } { 2 }$
(4) $\frac { 8 } { 3 }$

A data consists of $n$ observations: $x_1, x_2, \ldots, x_n$. If $\sum_{i=1}^{n}(x_i + 1)^2 = 9n$ and $\sum_{i=1}^{n}(x_i - 1)^2 = 5n$, then the standard deviation of this data is
(1) 5
(2) $\sqrt{7}$
(3) $\sqrt{5}$
(4) 2

The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11, then the correct variance is
(1) 3.99
(2) 4.01
(3) 4.02
(4) 3.98

Let $X = \{x \in N : 1 \leq x \leq 17\}$ and $Y = \{ax + b : x \in X$ and $a, b \in R, a > 0\}$. If mean and variance of elements of $Y$ are 17 and 216 respectively then $a + b$ is equal to
(1) 7
(2) $-7$
(3) $-27$
(4) 9

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