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Question (21)

QN0020072

The median of the data 5, 8, 12, 15, 18 is 12. If 20 is included, the new median becomes:

(A) 10
(B) 12
(C) 13.5
(D) 15

Correct Answer: C

Explanation:

New middle values are 12 and 15. Median = (12 + 15) ÷ 2 = 13.5.

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Question (22)

QN0020073

The median of 3, 6, 9, 12, 15 is 9. If 1 is included, the new median becomes:

(A) 6
(B) 7.5
(C) 9
(D) 10.5

Correct Answer: B

Explanation:

New middle values are 6 and 9. Median = (6 + 9) ÷ 2 = 7.5.

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Question (23)

QN0020074

The mean of 8 numbers is 25. What is their total sum?

(A) 25
(B) 100
(C) 200
(D) 225

Correct Answer: C

Explanation:

Sum = Mean × Number of values = 25 × 8 = 200.

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Question (24)

QN0020075

The mean of 15 values is 134. The sum of the values is:

(A) 149
(B) 2010
(C) 1340
(D) 1500

Correct Answer: B

Explanation:

Sum = 134 × 15 = 2010.

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Question (25)

QN0020076

The mean of 8, 13, 10, 4, 5, 20, y, 10 is 10.375. Find y.

(A) 10
(B) 12
(C) 13
(D) 15

Correct Answer: C

Explanation:

Total sum = 10.375 × 8 = 83. Known sum = 70. Therefore y = 13.

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Question (26)

QN0020077

The average weight of 10 players is 39.2 kg. Nine weights sum to 349 kg. The missing weight is:

(A) 41 kg
(B) 42 kg
(C) 43 kg
(D) 44 kg

Correct Answer: C

Explanation:

Total weight = 39.2 × 10 = 392 kg. Missing weight = 392 − 349 = 43 kg.

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Question (27)

QN0020078

An average harvest per tree is 25.6 for 15 trees. One tree’s count was recorded 3 more than actual. The correct average is:

(A) 25.2
(B) 25.4
(C) 25.6
(D) 25.8

Correct Answer: B

Explanation:

Actual total = (25.6 × 15) − 3 = 381. Average = 381 ÷ 15 = 25.4.

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Question (28)

QN0020079

Family sizes and frequencies are: 3(3), 4(11), 5(9), 6(7), 7(3), 8(1), 9(1), 10(1). The total number of families is:

(A) 34
(B) 35
(C) 36
(D) 37

Correct Answer: C

Explanation:

Total frequency = 3 + 11 + 9 + 7 + 3 + 1 + 1 + 1 = 36.

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Question (29)

QN0020080

Using the family-size data above, the mean family size is approximately:

(A) 5.00
(B) 5.22
(C) 5.50
(D) 6.22

Correct Answer: B

Explanation:

Weighted mean = 188 ÷ 36 ≈ 5.22.

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Question (30)

QN0020081

Using the same frequency table, the median family size is:

(A) 4
(B) 4.5
(C) 5
(D) 6

Correct Answer: C

Explanation:

The 18th and 19th observations lie in the value 5.

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Question (21)

The median of the data 5, 8, 12, 15, 18 is 12. If 20 is included, the new median becomes:

Explanation:

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Question (22)

The median of 3, 6, 9, 12, 15 is 9. If 1 is included, the new median becomes:

Explanation:

Link:

Question (23)

The mean of 8 numbers is 25. What is their total sum?

Explanation:

Link:

Question (24)

The mean of 15 values is 134. The sum of the values is:

Explanation:

Link:

Question (25)

The mean of 8, 13, 10, 4, 5, 20, y, 10 is 10.375. Find y.

Explanation:

Link:

Question (26)

The average weight of 10 players is 39.2 kg. Nine weights sum to 349 kg. The missing weight is:

Explanation:

Link:

Question (27)

An average harvest per tree is 25.6 for 15 trees. One tree’s count was recorded 3 more than actual. The correct average is:

Explanation:

Link:

Question (28)

Family sizes and frequencies are: 3(3), 4(11), 5(9), 6(7), 7(3), 8(1), 9(1), 10(1). The total number of families is:

Explanation:

Link:

Question (29)

Using the family-size data above, the mean family size is approximately:

Explanation:

Link:

Question (30)

Using the same frequency table, the median family size is:

Explanation:

Link:

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