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Statistics (Mean, Median, Mode) Questions and Answers

Important Instructions
21.

Find the weighted mean for the following data :

Value :1012151820
Weight :251247

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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Direction for [ Question No: 22 To 22 ] :

Consider the following information:

Time in Minutes that 16 Customers Waited in Line

Minutes56810131519
Customers3214231

22.

1.Number of customers who waited less than 15 minutes in line is 12The largest amount of time in minutes a customer waited in line is 19The least amount of time in minutes a customer waited in line is 10
2.Number of customers waited less than 15 minutes in line is 12The largest amount of time in minutes a customer waited in line is 19The least amount of time in minutes a customer waited in line is 5
3.Number of customers who waited less than 15 minutes in line is 15The largest amount of time in minutes a customer waited in line is 19The least amount of time in minutes a customer waited in line is 5

In how many of the above rows is the given information correctly tabulated?

Exam Year(s)

Haryana Civil Services - 2026, HPSC - 2026,

Answer: A

The information is correct in only one row.

From the data, 3 + 2 + 1 + 4 + 2 = 12 customers waited less than 15 minutes (the 5, 6, 8, 10 and 13 minute values). The longest wait is 19 minutes and the shortest is 5 minutes. The counts add to 16, as the title says.

  • Row 1 gives the least time as 10; it is 5. Wrong.
  • Row 2: 12, 19 and 5. All correct.
  • Row 3 gives 15 customers under 15 minutes, which counts the three customers who waited exactly 15; 'less than 15' excludes them. Wrong.

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23.

The median of the following data is :

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

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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24.

Find the median from the following data :

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

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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25.

Median of 12, 17, 5, 8, 13, 6, 9 is

Exam Year(s)

TNPSC Group 1 - 2017, TNPSC - 2017,

Answer: B

The median is 9.

Arrange the seven values in order: 5, 6, 8, 9, 12, 13, 17. With an odd number of values the median is the middle one, the 4th, which is 9.

8 and 12 are the values either side of the middle, and 17 is the largest value. Taking the middle of the list in its printed order (8) without sorting is the usual mistake.

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26.

Range and Range coefficient of the data -3, -2, -1, 0, 1, 2, 3 are respectively

Exam Year(s)

TNPSC Group 1 - 2017, TNPSC - 2017,

Answer: D

The range is 6 and the coefficient of range is infinite, so the answer is 6 and \(\infty\).

With the largest value \(L = 3\) and the smallest \(S = -3\):

\[ \begin{aligned} \text{Range} &= L - S = 3 - (-3) = 6 \\ \text{Coefficient} &= \frac{L - S}{L + S} = \frac{6}{0} \end{aligned} \]

Division by zero has no finite value, so the coefficient is taken as infinite, which is why this measure is meaningless for data centred on zero. Options giving the range as 0 forget the sign of -3, and '6 and 0' wrongly treats \(\frac{6}{0}\) as 0.

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27.

In a normal distribution, what percentage of the data lies within one Standard deviation from the Mean ?

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: A

About 68% of the data lie within one standard deviation of the mean.

By the empirical (68-95-99.7) rule for a normal distribution, about 68% of values lie within one standard deviation of the mean, 95% within two, and 99.7% within three.

So 95% and 99.7% are the two- and three-standard-deviation figures, and 50% is the share on either side of the mean.

The rule holds exactly only for normally distributed data; for other distributions Chebyshev's inequality gives weaker guarantees.

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28.

If a dataset has a skewness of 0, it indicates that

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: A

Zero skewness indicates that The dataset is symmetric.

Skewness measures asymmetry. A skewness of 0 means the distribution is balanced about its mean, with neither tail longer.

A normal distribution has zero skewness, but so do many other symmetric distributions, so zero skewness does not prove normality. Negative skewness means a longer left tail, and skewness says nothing directly about outliers.

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Answer: A

The z-score tells you how far a data point is from the Mean in terms of Standard deviations.

\(z = (x - \mu)/\sigma\). A z-score of 2 means the value is two standard deviations above the mean, and -1 one standard deviation below.

Percentages and likelihoods can be read from a z-table using the z-score, but the z-score itself is a distance in standard deviations, not a percentage or a probability, and not an average of deviations.

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30.

Which of the following statements can be inferred about the Mean, Median, and Mode of the normal distribution P(x) for a variable x ?

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: B

For a normal distribution MEAN = MEDIAN = MODE.

The normal curve is symmetric and has a single peak at its centre. Symmetry puts the mean and the median at the centre, and the single peak puts the mode there too, so all three coincide.

The options with an inequality describe skewed distributions, where the three measures separate.

In a positively skewed distribution the mean is pulled above the median, and in a negatively skewed one below it, which is why unequal measures signal skew.

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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