Statistics(11th Grade > Mathematics ) Questions and answers for exam Preparation

11th Grade > Mathematics

STATISTICS MCQs

Total Questions : 30 | Page 2 of 3 pages

Question 11.

If the range of discrete data of n observations is zero, then

all values of data are zero
all values of data are equal to standard deviation
all values of data are equal
the extreme values of data are different

Discuss Question

Answer: Option C. -> all values of data are equal
:
C

Let m be the least value of the discrete data and M be the maximum value of the discrete data.
Then range = M – m = 0 (hypothesis)
$∴$ M = m
$∴$ m $⩽$ each data item $⩽$ M = m
$\Rightarrow$ each data item = m or M i.e, all values of the data are equal.

Question 12.

The mean deviation of 2, 4, 6, 8, 10 about its mean is

Discuss Question

Answer: Option B. -> 2.4
:
B

M e a n (¯ x) = \frac{\sum x_{i}}{n} = \frac{2 + 4 + 6 + 8 + 10}{5} = 6 ∴ M . D = \frac{\sum | x_{i} - ¯ x |}{n} = \frac{| 2 - 6 | + | 4 - 6 | + | 6 - 6 | + | 8 - 6 | + | 10 - 6 |}{5} = \frac{4 + 2 + 0 + 2 + 4}{5} = \frac{12}{5} = 2.4

Question 13.

The coefficient of variation of a distribution is

$100 \times \frac{s . d}{m e a n}$
$100 \times \frac{m e a n}{s . d}$
$100 \times \frac{v a r i a n c e}{m e a n}$
$\frac{s . d}{m e a n}$

Discuss Question

Answer: Option A. ->

100 \times \frac{s . d}{m e a n}

:
A

C . V . i s d e f i n e d b y C . V = 100 \times \frac{s . d}{m e a n}

Question 14.

If a variable takes the discrete values $α + 4, α - \frac{7}{2}, α - \frac{5}{2}, α - 3, α - 2, α + \frac{1}{2}, α - \frac{1}{2}, α + 5 (α > 0)$ ,then the median is

$α - \frac{5}{4}$
$α - \frac{1}{2}$
$α - 2$
$α + \frac{5}{4}$

Discuss Question

Answer: Option A. ->

α - \frac{5}{4}

:
A
Arrange the data as:

α - \frac{7}{2}, α - 3, α - \frac{5}{2}, α - 2, α - \frac{1}{2}, α + \frac{1}{2}, α + 4, α + 5,

Median

= \frac{a - 2 + α - \frac{1}{2}}{2} = \frac{2 α - \frac{5}{2}}{2} = α - \frac{5}{4}

Question 15.

The variance of the data 2, 4, 6, 8, 10 is:

6
7
8
none of these

Discuss Question

Answer: Option C. -> 8
:
C

$¯ x = 6, \sum (x - ¯ x)^{2} = 40 ∴ v a r i a n c e = σ^{2} = \frac{40}{5} = 8$

Question 16.

If the mean of the number $27 + x, 31 + x, 89 + x, 107 + x, 156 + x$ is $82,$ then the mean of $130 + x, 126 + x, 68 + x, 50 + x, 1 + x$ is

Discuss Question

Answer: Option A. -> 75
:
A

Given,

82 = \frac{(27 + x) + (31 + x) + (89 + x) + (107 + x) + (156 + x)}{5}

\Rightarrow 82 \times 5 = 410 + 5 x \Rightarrow 410 - 410 = 5 x \Rightarrow x = 0

∴ Required mean is,

¯ ¯ ¯ x = \frac{130 + x + 126 + x + 68 + x + 50 + x + 1 + x}{5}

¯ ¯ ¯ x = \frac{375 + 5 x}{5} = \frac{375 + 0}{5} = \frac{375}{5} = 75.

Question 17.

The mean of n items is $¯ ¯ ¯ x .$ If the first term is increased by 1 second by 2 and so on, then new mean is

$¯ ¯ ¯ x + n$
$¯ ¯ ¯ x + \frac{n}{2}$
$¯ ¯ ¯ x + \frac{n + 1}{2}$
None of these

Discuss Question

Answer: Option C. ->

¯ ¯ ¯ x + \frac{n + 1}{2}

:
C

Let, $x_{1}, x_{2} . . . . . . . x_{n}$ be n items.
Then, $¯ ¯ ¯ x = \frac{1}{n} \sum x_{i}$
$L e t y_{1} = x_{1} + 1, y_{2} = x_{2} + 2, y_{3} = x_{3} + 3, . . . . ., y_{n} = x_{n} + n$
Then the mean of the new series is
$\frac{1}{n} \sum y_{i} = \frac{1}{n} n \sum i = 1 (x_{i} + i)$
$= \frac{1}{n} n \sum i = 1 x_{i} + \frac{1}{n} (1 + 2 + 3 + . . . . . . + n)$
$= ¯ ¯ ¯ x + \frac{1}{n} . \frac{n (n + 1)}{2} = ¯ ¯ ¯ x + \frac{n + 1}{2}$

Question 18.

If the algebraic sum of deviations of 20 observations from 30 is 20, then the mean of observations is

30
30.1
29
31

Discuss Question

Answer: Option D. -> 31
:
D

20 \sum i = 1 (x_{i} - 30) = 20 = 20 \sum i = 1 x_{i} - 20 \times 30 = 20

\Rightarrow 20 \sum i = 1 x_{i} = 620. M e a n = \frac{20 \sum i = 1}{20} = \frac{620}{20} = 31.

Question 19.

The mean and S.D. of the marks of 200 candidates were found to be 40 and 15 respectively. Later, it was discovered that a score of 40 was wrongly read as 50. The correct mean and S.D. respectively are

14.98, 39.95
39.95, 14.98
39.95, 224.5
None the these

Discuss Question

Answer: Option B. -> 39.95, 14.98
:
B

Corrected \sum x = 40 \times 200 - 50 + 40 = 7990

∴ Corrected ¯ ¯ ¯ x = \frac{7990}{200} = 39.95

\sum x^{2} = n [σ^{2} + {¯ ¯ ¯ x}^{2}] = 200 [15^{2} + 40^{2}] = 365000

Correct \sum x^{2} = 365000 - 2500 + 1600 = 364100

∴ Corrected σ = \sqrt{\frac{364100}{200} - (39.95)^{2}}

= \sqrt{(1820.5 - 1596)} = \sqrt{224.5} = 14.98.

Question 20.

Suppose a population A has 100 observations 101, 102,……200 and another population B has 100 observation 151, 152….., 250. If $V_{A}, V_{B}$ represent the variances of the two populations respectively, then $\frac{V_{A}}{V_{B}}$ is

$\frac{4}{9}$
$\frac{2}{3}$
1
$\frac{9}{4}$

Discuss Question

Answer: Option C. -> 1
:
C

Population B observations are 151, 152, ….., 250
Let y, be these observations
Shifting the origin $z_{i} = y_{i} - 50$
The values of these observations are 100, 101, ….., 200.
$∴ V_{A} = V_{B}$
( $∵$ Variance is independent of shifting the origin).