Free Objective Test 02 Practice Test - 11th and 12th

Question 1

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

SOLUTION

Solution : 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 2

If the median and mode of 4 observations is 4, 6 and the sum of squares of observations is 48. Then
standard deviation is

\sqrt{2}

B. 4

\sqrt{3}

\sqrt{5}

SOLUTION

Solution : C

Given that median = 4 and mode = 6. We know that
Mean - mode = 3 (Mean - Median)
Mean - 6 = 3 (Mean - 4)
$¯ x - 6 = 3 ¯ x - 12$
$¯ x = 3$
$σ^{2} = \frac{\sum x_{i}^{2}}{n} - {(\frac{\sum x^{i}}{n})}^{2} = \frac{48}{4} - (3)^{2} = 3$ $∴ σ = \sqrt{3}$

Question 3

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}

SOLUTION

Solution : 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 4

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

A. 4.2

B. 2.4

C. 3.4

D. 4.3

SOLUTION

Solution : 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 5

The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observations in the set is increased by 2, then the median of the new set

A. is increased by 2

B. is decreased by 2

C. is two times the original median

D. remains the same as that of the original set

SOLUTION

Solution : D

We know that median is found when we arrange the observations in ascending or descending order. And if the number of observations are odd then it is the $\frac{n + 1}{2} t h$ term.
Since on arranging 9 observations 5th will be the median.
On changing 4 of any side won't affect the median and it'll remain same.

Question 6

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}

SOLUTION

Solution : 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 7

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

A. 6

B. 7

C. 8

D. none of these

SOLUTION

Solution : C

$M e a n (¯ x) = \frac{\sum x_{i}}{n} = (\frac{(2 + 4 + 6 + 8 + 10)}{5} = 6$

$V a r i a n c e (σ^{2}) = \frac{\sum (x - ¯ x)^{2}}{n} = \frac{(2 - 6)^{2} + (4 - 6)^{2} + (6 - 6)^{2} + (8 - 6)^{2} + (10 - 6)^{2}}{5} = \frac{(16 + 4 + 0 + 4 + 16)}{5} = \frac{40}{5} = 8$

Question 8

The mean of a set of observation is $¯ ¯ ¯ x$ .If each observation is divided by $α α \neq 0$ , and then is increased by 10,
then the mean of the new set is

\frac{¯ ¯ ¯ x}{α}

\frac{¯ ¯ ¯ x + 10}{α}

\frac{¯ ¯ ¯ x + 10 α}{α}

α ¯ ¯ ¯ x + 10

SOLUTION

Solution : C

Let $x_{1}, x_{2} \dots \dots, x_{n}$ be n observations.
$Then ¯ ¯ ¯ x = \frac{1}{n} \sum x_{i}$
$let y_{i} = \frac{x_{i}}{α} + 10$
$¯ ¯ ¯ y = \frac{1}{n} n \sum i y_{i}$
$And, ¯ ¯ ¯ y = \frac{1}{n} [(n \sum i \frac{x_{i}}{α} + 10. n]$
$\Rightarrow ¯ ¯ ¯ y = \frac{1}{α} ¯ ¯ ¯ x + 10$
$= \frac{¯ ¯ ¯ x + 10 α}{α} .$

Question 9

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

A. 75

B. 157

C. 82

D. 80

SOLUTION

Solution : 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 10

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

A. 30

B. 30.1

C. 29

D. 31

SOLUTION

Solution : 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 11

The mean weight per student in a group of seven students is 55 kg if the individual weights of 6 students are 52, 58, 55, 53, 56 and 54; then weight of the seventh student is

A. 55 kg

B. 60 kg

C. 57 kg

D. 50 kg

SOLUTION

Solution : C

Total weight of 7 students is $= 55 \times 7 = 385 k g$
                Sum of weight of 6 students
                 $= 52 + 58 + 55 + 53 + 56 + 54 = 328 k g$
                 $∴$ Weight of seventh student $= 385 - 328 = 57 k g .$

Question 12

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

None of the above

$¯ ¯ ¯ x + \frac{n}{2}$

$¯ ¯ ¯ x + \frac{n + 1}{2}$

$¯ ¯ ¯ x + n$

SOLUTION

Solution : 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 13

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

A. 14.98, 39.95

B. 39.95, 14.98

C. 39.95, 224.5

D. None the these

SOLUTION

Solution : 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 14

If the coefficient of variation of data pertaining to goals scored by two teams A and B in a football season are 2.5 and 7.2 respectively, then which of the following statements i strue?

A. Team B is more consistent than team A.

B. Two teams are equally consistent.

C. Team A is more consistent than team B.

D. The s.ds. of team A and team B are

\frac{1}{25}, \frac{1}{72} .

SOLUTION

Solution : C

We know that, lesser is the coefficient of variation, the data is more consistent.

Since C.V. of the team A is 2.5, which is less than the C.V. of team B, we can conclude that team A is more Consistent.

Question 15

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

$\frac{4}{9}$

$\frac{2}{3}$

$\frac{9}{4}$

SOLUTION

Solution : 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).

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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