Let $\bar { x } , M$ and $\sigma ^ { 2 }$ be respectively the mean, mode and variance of $n$ observations $x _ { 1 } , x _ { 2 } , \ldots , x _ { n }$ and $d _ { i } = - x _ { i } - a , i = 1,2 , \ldots , n$, where $a$ is any number. Statement I: Variance of $d _ { 1 } , d _ { 2 } , \ldots , d _ { n }$ is $\sigma ^ { 2 }$. Statement II: Mean and mode of $d _ { 1 } , d _ { 2 } , \ldots , d _ { n }$ are $- \bar { x } - a$ and $- M - a$, respectively.
(1) Statement I and Statement II are both true
(2) Statement I and Statement II are both false
(3) Statement I is true and Statement II is false
(4) Statement I is false and Statement II is true

The variance of the first 50 even natural numbers is:
(1) 437
(2) $\frac { 437 } { 4 }$
(3) $\frac { 833 } { 4 }$
(4) 833

In a set of $2n$ distinct observations, each of the observation below the median of all the observations is increased by 5 and each of the remaining observations is decreased by 3. Then, the mean of the new set of observations:
(1) Increases by 2.
(2) Increase by 1.
(3) Decreases by 2.
(4) Decreases by 1.

The mean of a data set comprising of 16 observations is 16. If one of the observation value 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data is
(1) 14.0
(2) 16.8
(3) 16.0
(4) 15.8

The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is:
(1) 16.8
(2) 15.8
(3) 14.0
(4) 16.0

The variance of first 50 even natural numbers is:
(1) $833$
(2) $437$
(3) $\frac{833}{4}$
(4) $833$

$\lim _ { x \rightarrow 0 } \frac { ( 1 - \cos 2 x ) ^ { 2 } } { 2 x \tan x - x \tan 2 x }$ is
(1) 2
(2) $- \frac { 1 } { 2 }$
(3) $- 2$
(4) $\frac { 1 } { 2 }$

If the standard deviation of the numbers $2, 3, a$ and $11$ is $3.5$, then which of the following is true? (1) $3a^2 - 26a + 55 = 0$ (2) $3a^2 - 32a + 84 = 0$ (3) $3a^2 - 34a + 91 = 0$ (4) $3a^2 - 23a + 44 = 0$

If the standard deviation of the numbers $2, 3, a$ and $11$ is $3.5$, then which of the following is true?
(1) $3a^2 - 26a + 55 = 0$
(2) $3a^2 - 32a + 84 = 0$
(3) $3a^2 - 34a + 91 = 0$
(4) $3a^2 - 23a + 44 = 0$

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) 35
(2) 40
(3) 25
(4) 30

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 and variance for seven observations are 8 and 16 respectively. If 5 of the observations are $2, 4, 10, 12, 14$, then the product of the remaining two observations is
(1) 48
(2) 45
(3) 49
(4) 40

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