Statistics (Mean, Median, Mode) Questions and Answers

Q: Which of the following assertions are CORRECT?P: Adding 7 to each entry in a list adds 7 to the mean of the listQ: Adding 7 to each entry in a list adds 7 to the standard deviation of the listR: Doubling each entry in a list doubles the mean of the listS: Doubling each entry in a list leaves the standard deviation of the list unchanged

The correct assertions are P and R.P is true: adding a constant to every entry adds it to the mean.Q is false: adding a constant shifts every entry and the mean equally, so the deviations from the mean, and the standard deviation, are unchanged.R is true: doubling every entry doubles the mean.S is false: doubling every entry doubles every deviation, so the standard deviation doubles.Every other option includes Q or S, so P, R is the only fully correct choice.

Q: Marks3456789Frequency391171424The above frequency chart shows the frequency distribution of marks obtained by a set of students in an exam.From the data presented above, which one of the following is CORRECT?

The correct order is mode \(\gt\) median \(\gt\) mean.There are \(3 + 9 + 11 + 7 + 14 + 2 + 4 = 50\) students.Mode: the highest frequency, 14, is at 7 marks.Median: the 25th and 26th values. Cumulative frequencies are 3, 12, 23, 30, ..., so both fall at 6 marks: median 6.Mean: \((9 + 36 + 55 + 42 + 98 + 16 + 36)/50 = 292/50 = 5.84\).So 7 \(\gt\) 6 \(\gt\) 5.84. The other options put the mean above the median or the median above the mode, which these numbers contradict.

Q: Find the weighted mean for the following data :Value :1012151820Weight :251247

The weighted mean is about 15.7.Multiply each value by its weight, add, and divide by the total weight:\[ \begin{aligned} \textstyle\sum wx &= 20 + 60 + 180 + 72 + 140 = 472 \\ \textstyle\sum w &= 2 + 5 + 12 + 4 + 7 = 30 \end{aligned} \]\[ \bar{x}_{w} = 472/30 = 15.73 \]The simple mean of the five values is \(75/5 = 15\), which ignores the weights. 15.2, 15.6 and 14.3 come from arithmetic slips.

Q: The median of the following data is :17, 8, 9, 5, 26, 19, 13, 26, 29

The median is 17.Arrange the nine values in order:5, 8, 9, 13, 17, 19, 26, 26, 29With an odd number of values (9), the median is the middle one, the (9 + 1)/2 = 5th value, which is 17.13 and 19 are the values either side of the middle, and 18 would be the median only if there were an even number of values with 17 and 19 in the middle, which there are not.

Q: Find the median from the following data :36, 72, 49, 45, 60, 33, 52, 61 and 72

The median is 52.Arrange the nine values in order:33, 36, 45, 49, 52, 60, 61, 72, 72With an odd number of values (9), the median is the middle one, the 5th, which is 52.49 is the 4th value and 60 the 6th, so choosing either means miscounting the position. 61 is what you get by taking the 5th value of the unsorted list without arranging it first.

Important Instructions

81.

Which of the following assertions are CORRECT?

P: Adding 7 to each entry in a list adds 7 to the mean of the list
Q: Adding 7 to each entry in a list adds 7 to the standard deviation of the list
R: Doubling each entry in a list doubles the mean of the list
S: Doubling each entry in a list leaves the standard deviation of the list unchanged

A.P, Q
B.Q, R
C.P, R
D.R, S

Exam Year(s)

GATE ME - 2012,

Answer: C

The correct assertions are P and R.

P is true: adding a constant to every entry adds it to the mean.
Q is false: adding a constant shifts every entry and the mean equally, so the deviations from the mean, and the standard deviation, are unchanged.
R is true: doubling every entry doubles the mean.
S is false: doubling every entry doubles every deviation, so the standard deviation doubles.

Every other option includes Q or S, so P, R is the only fully correct choice.

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82.

Marks	3	4	5	6	7	8	9
Frequency	3	9	11	7	14	2	4

The above frequency chart shows the frequency distribution of marks obtained by a set of students in an exam.

From the data presented above, which one of the following is CORRECT?

A.mean \(\gt\) mode \(\gt\) median
B.mode \(\gt\) median \(\gt\) mean
C.mode \(\gt\) mean \(\gt\) median
D.median \(\gt\) mode \(\gt\) mean

Exam Year(s)

GATE Physics - 2022,

Answer: B

The correct order is mode \(\gt\) median \(\gt\) mean.

There are \(3 + 9 + 11 + 7 + 14 + 2 + 4 = 50\) students.

Mode: the highest frequency, 14, is at 7 marks.
Median: the 25th and 26th values. Cumulative frequencies are 3, 12, 23, 30, ..., so both fall at 6 marks: median 6.
Mean: \((9 + 36 + 55 + 42 + 98 + 16 + 36)/50 = 292/50 = 5.84\).

So 7 \(\gt\) 6 \(\gt\) 5.84. The other options put the mean above the median or the median above the mode, which these numbers contradict.

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83.

Find the weighted mean for the following data :

Value :	10	12	15	18	20
Weight :	2	5	12	4	7

A.15.2
B.15.6
C.15.7
D.14.3

Exam Year(s)

ISRO Assistant - 2023,

Answer: C

The weighted mean is about 15.7.

Multiply each value by its weight, add, and divide by the total weight:

\[ \begin{aligned} \textstyle\sum wx &= 20 + 60 + 180 + 72 + 140 = 472 \\ \textstyle\sum w &= 2 + 5 + 12 + 4 + 7 = 30 \end{aligned} \]

\[ \bar{x}_{w} = 472/30 = 15.73 \]

The simple mean of the five values is \(75/5 = 15\), which ignores the weights. 15.2, 15.6 and 14.3 come from arithmetic slips.

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84.

The median of the following data is :

17, 8, 9, 5, 26, 19, 13, 26, 29

A.13
B.17
C.18
D.19

Exam Year(s)

MPPSC State Service - 2024, MPPSC Forest Service - 2024,

Answer: B

The median is 17.

Arrange the nine values in order:

5, 8, 9, 13, 17, 19, 26, 26, 29

With an odd number of values (9), the median is the middle one, the (9 + 1)/2 = 5th value, which is 17.

13 and 19 are the values either side of the middle, and 18 would be the median only if there were an even number of values with 17 and 19 in the middle, which there are not.

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85.

Find the median from the following data :

36, 72, 49, 45, 60, 33, 52, 61 and 72

A.52
B.60
C.49
D.61

Exam Year(s)

MPPSC State Service - 2023, MPPSC Forest Service - 2023,

Answer: A

The median is 52.

Arrange the nine values in order:

33, 36, 45, 49, 52, 60, 61, 72, 72

With an odd number of values (9), the median is the middle one, the 5th, which is 52.

49 is the 4th value and 60 the 6th, so choosing either means miscounting the position. 61 is what you get by taking the 5th value of the unsorted list without arranging it first.

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Tags: Statistics Mean, Median, Mode Questions || Data Statistics MCQ || Mean and Median GK || Mode Problems Quantitative Aptitude || Statistics Quantitative Aptitude MCQs

Statistics (Mean, Median, Mode) Questions and Answers

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Other Subjects

Tags: Statistics Mean, Median, Mode Questions || Data Statistics MCQ || Mean and Median GK || Mode Problems Quantitative Aptitude || Statistics Quantitative Aptitude MCQs

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