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Question (11)

QN0037081

If the mean of a distribution is 24 and the median is 26, the mode is:

(A) 26
(B) 28
(C) 30
(D) 32

Correct Answer: B

Explanation:

Mode = 3(26) − 2(24) = 78 − 48 = 30.

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Question (12)

QN0037082

For the data 5, 7, 7, 8, 9, the mode is:

(A) 5
(B) 7
(C) 8
(D) 9

Correct Answer: B

Explanation:

7 occurs most frequently.

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Question (13)

QN0037083

For a highly skewed distribution, which measure is generally more suitable for describing the central position?

(A) Median
(B) Mean only
(C) Mode only
(D) Class size

Correct Answer: A

Explanation:

Median is less affected by extreme values and is useful for skewed distributions.

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Question (14)

QN0037084

If the mean is 20 and the median is 22, the mode according to the empirical relation is:

(A) 24
(B) 25
(C) 26
(D) 28

Correct Answer: C

Explanation:

Mode = 3(22) − 2(20) = 66 − 40 = 26.

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Question (15)

QN0037085

In the assumed mean method, the deviation dᵢ is calculated as:

(A) xᵢ + A
(B) xᵢ − A
(C) A − fᵢ
(D) xᵢ/fᵢ

Correct Answer: B

Explanation:

In the assumed mean method, dᵢ = xᵢ − A.

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Question (16)

QN0037086

The class marks of three classes are 5, 15 and 25 with frequencies 2, 4 and 4. The mean is:

(A) 15
(B) 17
(C) 18
(D) 20

Correct Answer: B

Explanation:

Mean = (2×5 + 4×15 + 4×25)/10 = 170/10 = 17.

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Question (17)

QN0037087

The class marks are 10, 20, 30 and 40 with frequencies 2, 3, 4 and 1. The mean is:

(A) .
(B) 25
(C) 26
(D) 27

Correct Answer: C

Explanation:

Mean = (20 + 60 + 120 + 40)/10 = 240/10 = 24. Correct answer is (a).

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Question (18)

QN0037088

For the class intervals 0–10, 10–20, 20–30, the class marks are respectively:

(A) 0, 10, 20
(B) 5, 15, 25
(C) 10, 20, 30
(D) 5, 10, 15

Correct Answer: B

Explanation:

Class mark = (lower limit + upper limit)/2.

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Question (19)

QN0037089

If A = 25 and xᵢ = 30, then the deviation dᵢ is:

(A) 3
(B) 5
(C) 10
(D) 15

Correct Answer: B

Explanation:

dᵢ = xᵢ − A = 30 − 25 = 5.

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Question (20)

QN0037090

If A = 30 and the deviations are −10, −5, 0, 5 and 10 with frequencies 2, 3, 4, 3 and 2, the value of Σfᵢdᵢ is:

(A) 0
(B) 5
(C) 10
(D) 15

Correct Answer: A

Explanation:

Σfd = −20 − 15 + 0 + 15 + 20 = 0.

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Question (11)

If the mean of a distribution is 24 and the median is 26, the mode is:

Explanation:

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Question (12)

For the data 5, 7, 7, 8, 9, the mode is:

Explanation:

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Question (13)

For a highly skewed distribution, which measure is generally more suitable for describing the central position?

Explanation:

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Question (14)

If the mean is 20 and the median is 22, the mode according to the empirical relation is:

Explanation:

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Question (15)

In the assumed mean method, the deviation dᵢ is calculated as:

Explanation:

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Question (16)

The class marks of three classes are 5, 15 and 25 with frequencies 2, 4 and 4. The mean is:

Explanation:

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Question (17)

The class marks are 10, 20, 30 and 40 with frequencies 2, 3, 4 and 1. The mean is:

Explanation:

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Question (18)

For the class intervals 0–10, 10–20, 20–30, the class marks are respectively:

Explanation:

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Question (19)

If A = 25 and xᵢ = 30, then the deviation dᵢ is:

Explanation:

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Question (20)

If A = 30 and the deviations are −10, −5, 0, 5 and 10 with frequencies 2, 3, 4, 3 and 2, the value of Σfᵢdᵢ is:

Explanation:

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