Statistics And Collection Of Statistical Data(12th Grade > Mathematics ) Questions and answers for exam Preparation

12th Grade > Mathematics

STATISTICS AND COLLECTION OF STATISTICAL DATA MCQs

Statistics, Introduction To Statistics, Collection Of Data (11th And 12th Grade)

Total Questions : 105 | Page 10 of 11 pages

Question 91. 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 92. 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 93. 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+n2
¯¯¯x+n+12
None of these

Discuss Question

Answer: Option C. -> ¯¯¯x+n+12
:
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 94. If the mean of the number

27 + x, 31 + x, 89 + x, 107 + x, 156 + x

82,

then the mean of

130 + x, 126 + x, 68 + x, 50 + x, 1 + x

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 95. 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 96. 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}}

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).

Question 97. Mean deviation of the series a, a + d, a + 2d, a + 2nd from its mean is

(n+1)d(2n+1)
(nd)(2n+1)
n(n+1)d(2n+1)
2(n+1)dn(n+1)

Discuss Question

Answer: Option C. -> n(n+1)d(2n+1)
:
C

¯ x = \frac{\frac{2 n + 1}{2} (a + a + 2 n d))}{(2 n + 1)} = a + n d \sum | x - ¯ x | = 2 d (1 + 2 + . . . . . + n) = n (n + 1) d ∴ M . D = \frac{n (n + 1) d}{(2 n + 1)}

Question 98. The mean of 10 numbers is 12.5; the mean of the first six is 15 and the last five is 10. The sixth number is

15
12
18
none of these

Discuss Question

Answer: Option A. -> 15
:
A
Let the mean of the last four be

A_{2}

.
Then by the formula for combined mean,

12.5 = \frac{6 \times 15 + 4 \times A_{2}}{6 + 4}; or 125=90+4 A_{2};

∴ A_{2} = \frac{35}{4}

Let the sixth number = x; then taking the sixth number as a collection, the combined mean of this collection and the collection of the last five is 10, by question.

∴

By definition of combined mean

10 = \frac{1 \times x + 4 \times \frac{35}{4}}{1 + 4}; 50 = x + 35; ∴ x = 15

∴

sixth number = 15

Question 99. If the coefficient of variation and standard deviation of a distribution are 2 and 0.4 respectively, then mean of the distribution is

Discuss Question

Answer: Option B. -> 20
:
B

C . V = 2; σ = 0.4 (g i v e n) C . V = 100 \times \frac{s . d}{m e a n} 2 = \frac{100 \times 0.4}{m e a n} M e a n = \frac{40}{2} = 20

Question 100. The A.M. of a set of 50 numbers is 38. If two numbers of the set, namely 55 and 45 are discarded, the A.M. of the remaining set of numbers is