Trimmed mean excludes

a percentage of extreme observations from both tails

the average of the first and third quartiles (Q1+Q3)/2

Measures of Central Tendency MCQs 2026

60 questions with detailed answers · 21 from past papers · 6 quiz batches available

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Page 1 of 1 Questions 1–10 of 60

Q1 hard

The mean of grouped data using midpoints is Σ(f·m) / Σf. For classes 0–9 (f=2), 10–19 (f=3) with midpoints 4.5 and 14.5, mean is
1. A 9.5
2. B 11
3. C 10.5
4. D 8.5
💡 Explanation:
(2×4.5+3×14.5)/5 = (9+43.5)/5 = 52.5/5 = 10.5.
Q2 hard

Trimmed mean excludes
1. A only the mode
2. B only the median
3. C a percentage of extreme observations from both tails
4. D only middle values
💡 Explanation:
Trimming reduces sensitivity to outliers.
Q3 hard

Midhinge is
1. A the same as the median always
2. B the average of the first and third quartiles (Q1+Q3)/2
3. C the product of Q1 and Q3
4. D the range divided by 2 only
💡 Explanation:
Midhinge = (Q1+Q3)/2.
Q4 easy

For symmetric unimodal data, mean, median and mode are
1. A always zero
2. B always undefined
3. C approximately equal
4. D median always twice the mean
💡 Explanation:
Symmetry places centre at the same point.
Q5 medium

If each observation is multiplied by 3, the new mean becomes
1. A unchanged
2. B 3 times the original mean
3. C increased by 3 only
4. D divided by 3
💡 Explanation:
Linear transformation: new mean = 3×old mean.
Q6 medium

If 5 is added to every observation, the new mean becomes
1. A original mean minus 5
2. B unchanged
3. C multiplied by 5
4. D original mean plus 5
💡 Explanation:
Adding a constant shifts the mean by that constant.
Q7 hard

The best average for open-ended grouped classes when midpoints are uncertain is often
1. A median from cumulative frequency
2. B mode from highest frequency class only
3. C range divided by 2
4. D harmonic mean of limits
💡 Explanation:
Median can be located from ogive without assuming extreme midpoints.
Q8 hard

Step deviation method computes mean using
1. A only geometric products
2. B only harmonic reciprocals
3. C only ranks without values
4. D assumed mean and coded deviations di = (xi − A)/h
💡 Explanation:
x̄ = A + (Σf·d)/Σf × h simplifies calculation.
Q9 easy

If AM of a dataset is 12 and each value is doubled, new AM is
1. A 24
2. B 12
3. C 14
4. D 6
💡 Explanation:
Doubling all values doubles the mean.
Q10 medium

A distribution with mode 10 and median 12 suggests positive skew if mean is
1. A less than 10
2. B exactly 10
3. C equal to median always
4. D greater than 12
💡 Explanation:
Positive skew: mean pulled above median.
Q11 medium

Average of first n natural numbers 1,2,...,n is
1. A n/2 only without +1
2. B n²
3. C (n−1)/2
4. D n(n+1)/(2n) which simplifies to (n+1)/2
💡 Explanation:
Sum = n(n+1)/2; divide by n gives (n+1)/2.
Q12 easy

Mean of 5, 5, 5, 5 is
1. A 0
2. B 20
3. C 5
4. D 25
💡 Explanation:
All equal values: mean equals that value.
Q13 easy

When all values are distinct and n is even, median is
1. A the arithmetic mean of the two middle values
2. B the larger middle value only
3. C the smaller middle value only
4. D the most frequent value
💡 Explanation:
Average of middle pair ensures even-n definition.
Q14 medium

Adding a value equal to the current mean will
1. A double the mean
2. B reduce the mean to zero
3. C always increase the median
4. D leave the mean unchanged
💡 Explanation:
Balance point unchanged when adding value at mean.
Q15 easy

For heavily skewed income data, the most representative central measure is usually
1. A mean
2. B midrange
3. C range
4. D median
💡 Explanation:
Median resists extreme high incomes.
Q16 medium

Mode can be determined graphically from
1. A a scatter plot of two variables
2. B a histogram as the midpoint of the modal class
3. C only a boxplot whisker
4. D only cumulative percentage at 50%
💡 Explanation:
Highest bar identifies modal class; midpoint estimates mode.
Q17 easy

If mean of 6 numbers is 15, their total is
1. A 15
2. B 75
3. C 90
4. D 6
💡 Explanation:
Total = 15 × 6 = 90.
Q18 hard

The average growth rate over two periods of 10% and 20% is best found using
1. A simple arithmetic mean of 10 and 20 only
2. B geometric mean of growth factors 1.10 and 1.20
3. C harmonic mean of 10 and 20
4. D median of 10 and 20
💡 Explanation:
GM of (1.10, 1.20) gives consistent compound growth average.
Q19 easy

CSS/PPSC often test
1. A only matrix algebra
2. B only complex integration
3. C only organic nomenclature
4. D mean, median, mode and quartiles on small datasets
💡 Explanation:
Competitive items use quick computation and comparison.
Q20 easy

When class frequencies are 5, 12, 8, highest frequency class is the
1. A median class always
2. B first class always
3. C modal class
4. D class with lowest midpoint
💡 Explanation:
Maximum frequency identifies the modal class.
Q21 medium

For dataset 1, 2, 2, 3, 100 the median is
1. A 1
2. B 3
3. C 2
4. D 21.6
💡 Explanation:
Ordered middle value is 2; median resists outlier 100.
Q22 medium

For dataset 1, 2, 2, 3, 100 the mean is
1. A 21.6
2. B 2
3. C 3
4. D 20
💡 Explanation:
Sum = 108; n = 5; mean = 108/5 = 21.6.
Q23 easy

Interquartile range is
1. A Q2 − Q1
2. B Q3 + Q1
3. C Range of entire data
4. D Q3 − Q1
💡 Explanation:
IQR covers middle 50% spread.
Q24 medium

If every observation increases by 10%, the new mean is
1. A unchanged
2. B increased by 10 units only
3. C 110% of the old mean
4. D decreased by 10%
💡 Explanation:
Proportional increase multiplies mean by 1.10.
Q25 medium

The median can be obtained from a cumulative frequency curve at
1. A total frequency N always
2. B zero frequency
3. C N/2 (or (N+1)/2 depending on convention)
4. D maximum frequency only
💡 Explanation:
Cumulative 50% point locates median.
Q26 easy

Average of 8 and 12 is
1. A 8
2. B 12
3. C 20
4. D 10
💡 Explanation:
(8+12)/2 = 10.
Q27 hard

If HM = 4 for two equal numbers, each number is
1. A 2
2. B 8
3. C 4
4. D 16
💡 Explanation:
When all values equal k, HM = GM = AM = k.
Q28 hard

Mean deviation is minimum when measured from
1. A always the minimum value
2. B the median (among all choices of reference point)
3. C always zero
4. D always the maximum
💡 Explanation:
Mean absolute deviation is minimized at the median.
Q29 easy

For percentages 40%, 60% the arithmetic mean is
1. A 48%
2. B 52%
3. C 50%
4. D 2400%
💡 Explanation:
(40+60)/2 = 50%.
Q30 easy

Mean of grouped data with Σf = 50 and Σ(f·m) = 1500 is
1. A 30
2. B 25
3. C 35
4. D 30000
💡 Explanation:
1500/50 = 30.
Q31 Past Paper · PPSC/FPSC/CSS easy

The median of 2, 4, 6, 8 is
1. A 4
2. B 6
3. C 5.5
4. D 5
💡 Explanation:
Even n: average of 4 and 6 = 5.
Q32 medium

The 90th percentile means
1. A about 90% of observations lie at or below that value
2. B 10% lie below
3. C 50% lie below
4. D exactly 90 observations
💡 Explanation:
Percentile p indicates position in ordered data.
Q33 medium

The sum of deviations from the arithmetic mean is always
1. A one
2. B equal to the variance
3. C equal to the median
4. D zero
💡 Explanation:
Σ(x − x̄) = 0 for any dataset.
Q34 easy

If two observations have mean 10, their sum is
1. A 10
2. B 20
3. C 5
4. D 100
💡 Explanation:
Sum = mean × n = 10 × 2 = 20.
Q35 Past Paper · PPSC/FPSC/CSS easy

The mode of 2, 3, 3, 5, 7 is
1. A 2
2. B 3
3. C 5
4. D 7
💡 Explanation:
3 occurs most frequently.
Q36 Past Paper · PPSC/FPSC/CSS easy

The arithmetic mean of 4, 6, 8, 10 is
1. A 7
2. B 6
3. C 8
4. D 28
💡 Explanation:
Sum = 28; n = 4; mean = 28/4 = 7.
Q37 Past Paper · PPSC/FPSC/CSS easy

A measure of central tendency describes
1. A the typical or central value of a dataset
2. B only the spread of data
3. C only the shape of tails only
4. D only sampling error
💡 Explanation:
Central measures summarize where data cluster.
Q38 Past Paper · PPSC/FPSC/CSS medium

Geometric mean is especially suitable for
1. A averaging rates of change and ratios
2. B nominal categorical labels
3. C simple addition of temperatures in °C
4. D unordered survey responses
💡 Explanation:
GM applies to multiplicative processes like growth rates.
Q39 Past Paper · PPSC/FPSC/CSS medium

The harmonic mean of 2 and 6 is
1. A 4
2. B 3
3. C 2
4. D 3.5
💡 Explanation:
HM = 2/(1/2+1/6) = 2/(4/6) = 3.
Q40 Past Paper · PPSC/FPSC/CSS hard

Harmonic mean is used when averaging
1. A plain counts of children
2. B nominal categories
3. C ranks without values
4. D rates involving reciprocals such as speeds over equal distances
💡 Explanation:
HM weights small values more; useful for average speed there and back.
Q41 Past Paper · PPSC/FPSC/CSS hard

For positive numbers, generally
1. A AM ≤ GM ≤ HM
2. B GM ≤ HM ≤ AM
3. C HM ≤ GM ≤ AM
4. D AM ≤ HM ≤ GM
💡 Explanation:
Equality holds only when all values are equal.
Q42 Past Paper · PPSC/FPSC/CSS easy

Quartiles divide ordered data into
1. A four equal parts
2. B two parts only
3. C ten parts only
4. D hundred parts only
💡 Explanation:
Q1, Q2 (median), Q3 are quartile positions.
Q43 Past Paper · PPSC/FPSC/CSS easy

For 2, 4, 6, 8, 10, 12, 14 the median Q2 is
1. A 6
2. B 8
3. C 10
4. D 7
💡 Explanation:
Middle of seven ordered values is 8.
Q44 Past Paper · PPSC/FPSC/CSS medium

For 2, 4, 6, 8, 10, 12, 14 the first quartile Q1 is
1. A 2
2. B 4
3. C 6
4. D 5
💡 Explanation:
Lower half (2,4,6) has median 4.
Q45 Past Paper · PPSC/FPSC/CSS medium

For 2, 4, 6, 8, 10, 12, 14 the third quartile Q3 is
1. A 12
2. B 10
3. C 14
4. D 11
💡 Explanation:
Upper half (10,12,14) has median 12.
Q46 Past Paper · PPSC/FPSC/CSS hard

The empirical relation among mean, median and mode for moderately skewed data is
1. A Mean = Mode always
2. B Median − Mode ≈ 3(Mean − Mode)
3. C Mean − Mode ≈ 3(Mean − Median)
4. D Mean + Mode ≈ 3 Median
💡 Explanation:
Pearson's approximation links the three averages.
Q47 Past Paper · PPSC/FPSC/CSS medium

In a positively skewed distribution, typically
1. A Mean < Median < Mode
2. B Mean = Median = Mode always
3. C Mode > Mean > Median
4. D Mean > Median > Mode
💡 Explanation:
Right tail pulls the mean upward.
Q48 Past Paper · PPSC/FPSC/CSS medium

In a negatively skewed distribution, typically
1. A Mean > Median > Mode
2. B Mean = Median always
3. C Mean < Median < Mode
4. D Median < Mode < Mean
💡 Explanation:
Left tail pulls the mean downward.
Q49 Past Paper · PPSC/FPSC/CSS easy

The mean is unduly affected by
1. A only the most frequent value
2. B only the middle order statistic
3. C only class width
4. D extreme values (outliers)
💡 Explanation:
Outliers shift the arithmetic mean.
Q50 Past Paper · PPSC/FPSC/CSS easy

The median is preferred to the mean when
1. A the distribution is skewed or has outliers
2. B data are perfectly symmetric always
3. C all values are equal
4. D only nominal data are recorded
💡 Explanation:
Median is resistant to extreme observations.
Q51 easy

The mode is the only measure of central tendency that can be used with
1. A ratio data only
2. B interval data only
3. C nominal data
4. D only continuous data
💡 Explanation:
Categories can have a most frequent label.
Q52 Past Paper · PPSC/FPSC/CSS easy

The median of 3, 7, 9, 11, 15 is
1. A 7
2. B 9
3. C 11
4. D 8
💡 Explanation:
Ordered odd n: middle value is 9.
Q53 hard

Combined mean of two groups with n1=10, mean1=20 and n2=20, mean2=30 is
1. A 25
2. B 27
3. C 28
4. D 26.67 (approximately)
💡 Explanation:
Combined = (10×20+20×30)/30 = 800/30 ≈ 26.67.
Q54 easy

GM of two numbers 4 and 16 is
1. A 10
2. B 8
3. C 6
4. D 12
💡 Explanation:
√(4×16) = √64 = 8.
Q55 medium

For equal values k repeated n times, AM, GM and HM all equal
1. A 0
2. B k
3. C n
4. D k/n
💡 Explanation:
No variation: all averages coincide at k.
Q56 medium

Quartile deviation is
1. A Q3 − Q1 without dividing
2. B (Q3 − Q1)/2
3. C Q3 + Q1
4. D Range/4 always
💡 Explanation:
Semi-interquartile range measures half the IQR.
Q57 Past Paper · PPSC/FPSC/CSS medium

The geometric mean of 1, 4, 16 is
1. A 4
2. B 7
3. C 5
4. D 6.33
💡 Explanation:
GM = (1×4×16)^(1/3) = 64^(1/3) = 4.
Q58 Past Paper · PPSC/FPSC/CSS medium

The weighted arithmetic mean of values 10 and 20 with weights 2 and 3 is
1. A 15
2. B 16
3. C 17
4. D 14
💡 Explanation:
(10×2+20×3)/(2+3) = 80/5 = 16.
Q59 hard

HM of 4 and 16 is
1. A 6.4
2. B 8
3. C 10
4. D 5
💡 Explanation:
HM = 2/(1/4+1/16) = 2/(5/16) = 32/5 = 6.4.
Q60 Past Paper · PPSC/FPSC/CSS medium

A dataset with two equal highest frequencies is called
1. A bimodal
2. B unimodal
3. C trimodal
4. D nonmodal
💡 Explanation:
Two modes indicate bimodality.

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