APPSC Paper I · Assistant Statistical Officer Paper II · Topic 7
7 APPSC Paper I Measures of Central Tendency questions with answers
26 questions from APPSC ASO 2025 Paper II question paper with answers and APPSC ASO Paper II Mock 1 with answers. Try each one, then tap Show answer.
Previous-paper questions are as in the original paper. Medha AI wrote the answers and explanations.
7.1Arithmetic, geometric & harmonic meansNotes
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Given below are two statements labelled as Assertion (A) and Reasoning (R).
Assertion (A): Mean is highly affected by extreme observations.
Reasoning (R): Mean does not utilize all the observations.
In the light of the above, choose the correct option.
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Answer C. A is true but R is not true.
The arithmetic mean is pulled by extreme values precisely because it uses every observation; so A is true and R (that it does not use all observations) is false.
Source: APPSC Assistant Statistical Officer 2025, Paper II (Notification 12/2024), Q70, final key
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If 5, 8, 6 and 2 occur with frequencies 3, 2, 4 and 1, then arithmetic mean is equal to:
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Answer A. 5.7
Mean = Σfx / Σf = (15 + 16 + 24 + 2) / 10 = 57 / 10 = 5.7.
Source: APPSC Assistant Statistical Officer 2025, Paper II (Notification 12/2024), Q76, final key
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If a typist in an office can type invoices at a rate of 50 per hour, statements at a rate of 40 per hour and reminders at a rate of 80 per hour and she types equal number of each type of documents, then the average output of the typist (per hour) is approximately equal to?
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Answer B. 52
For equal numbers of documents typed at different rates, use the harmonic mean: 3 / (1/50 + 1/40 + 1/80) = 3 / 0.0575 ≈ 52.
Source: APPSC Assistant Statistical Officer 2025, Paper II (Notification 12/2024), Q85, final key
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For the following set of observations
2,4,5,6,7,8,9, 9,9, 10,11
Match the following answers:
Column 1 Column 2 1. Mean I. 5.76 2. Mode II. 8 3. Median III. 9 4. Harmonic Mean IV. 7.27 Choose the correct option.
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Answer B. 1-IV; 2-III; 3- II; 4-I
n = 11, sum = 80, so mean = 7.27; 9 occurs most often, so mode = 9; the 6th value is 8, so median = 8; HM = 11 / Σ(1/x) = 11 / 1.909 ≈ 5.76.
Source: APPSC Assistant Statistical Officer 2025, Paper II (Notification 12/2024), Q97, final key
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Which of the following is not true about central tendency?
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Answer B. Indicates variability in a data set
A measure of central tendency is a single typical value around which data cluster; variability (spread) is measured by dispersion, not central tendency.
Source: APPSC Assistant Statistical Officer 2025, Paper II (Notification 12/2024), Q103, final key
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Statement 1: The mean of first 10 natural numbers is 10.
Statement 2: If 3 is subtracted from original observations, then mean does not change.
Statement 3: If original observations are multiplied by a constant, then median also gets multiplied by same constant.
On the basis of the above statements, which of the following statement/s is/are true?
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Answer C. Only 3 is true
The mean of 1 to 10 is 5.5, and subtracting 3 from every value lowers the mean by 3; multiplying all values by a constant multiplies the median by the same constant.
Source: APPSC Assistant Statistical Officer 2025, Paper II (Notification 12/2024), Q115, final key
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For 6 bags of rice, the mean weight per bag is 119 kg. The weights of 5 bags are 115 kg, 109 kg, 129 kg, 117 kg and 114 kg. The weight (in kg) of the sixth bag is equal to:
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Answer D. 130
Total = 6 x 119 = 714; the five bags weigh 584 kg, so the sixth weighs 714 - 584 = 130 kg.
Source: APPSC Assistant Statistical Officer 2025, Paper II (Notification 12/2024), Q131, final key
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A shop sold three grades of tea priced at ₹200, ₹250 and ₹300 per kg. The quantities sold were 50 kg, 30 kg and 20 kg respectively. What is the weighted average price per kg?
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Answer B. ₹235
Weighted mean = Σwx / Σw = (10,000 + 7,500 + 6,000) / 100 = ₹235.
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
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The mean of 10 observations is 20. Match the change made to every observation (List I) with the new mean (List II):
List I:
A. Each increased by 5
B. Each multiplied by 3
C. Each divided by 4
D. Each decreased by 20
List II:
1. 0
2. 5
3. 25
4. 60
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Answer C. A-3, B-4, C-2, D-1
A: 20 + 5 = 25; B: 3 × 20 = 60; C: 20 / 4 = 5; D: 20 − 20 = 0.
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
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The arithmetic mean of two positive numbers is 10 and their harmonic mean is 6.4. What are the two numbers?
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Answer B. 4 and 16
For two numbers GM² = AM × HM = 10 × 6.4 = 64, so GM = 8 and the product is 64. The sum is 2 × 10 = 20. Numbers with sum 20 and product 64 are 4 and 16 (check: HM = 2 × 64 / 20 = 6.4).
Source: Standard method; answer and every option computed and checked in Python
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The mean of 20 observations was found to be 15. Later it was found that one observation, 52, had been recorded as 25, and that another observation, 30, should not have been included at all. What is the correct mean of the remaining observations?
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Answer B. 15.63
Correct total = 300 − 25 + 52 − 30 = 297 for 19 observations, so the mean = 297 / 19 ≈ 15.63.
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
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Consider the following statements for positive observations:
1. If the observations are not all equal, AM > GM > HM.
2. For any two positive numbers, GM² = AM × HM.
3. For any three positive numbers, GM² = AM × HM.
Which of the statements given above is/are correct?
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Answer B. 1 and 2 only
AM ≥ GM ≥ HM with equality only when all values are equal. For two numbers a, b: AM × HM = ((a + b)/2) × (2ab/(a + b)) = ab = GM². This does not hold in general for three numbers: for 1, 2, 3, AM × HM = 2 × 18/11 ≈ 3.273 while GM² = 6^(2/3) ≈ 3.302.
Source: Standard method; answer and every option computed and checked in Python
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The harmonic mean of a group of 20 observations is 8 and that of another group of 30 observations is 15. What is the harmonic mean of all 50 observations?
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Answer D. 11.11
Σ(1/x) for the groups = 20/8 + 30/15 = 2.5 + 2 = 4.5, so the combined HM = 50 / 4.5 ≈ 11.11.
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
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For two positive numbers, the arithmetic mean is 25 and the harmonic mean is 16. What is their geometric mean?
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Answer B. 20
For two positive numbers GM² = AM × HM = 25 × 16 = 400, so GM = 20.
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
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If every weight in a weighted arithmetic mean is multiplied by 5, the weighted mean:
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Answer A. remains unchanged
x̄w = Σwx / Σw; multiplying every w by 5 multiplies both numerator and denominator by 5, which cancel.
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
7.1Median, mode, quartiles, deciles & percentilesNotes
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The numbers of ATM transactions per day recorded at 10 locations in a city were 35, 49, 225, 50, 30, 65, 40, 55, 52 and 76. The median of the numbers of transactions is:
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Answer B. 51
Sorted: 30, 35, 40, 49, 50, 52, 55, 65, 76, 225. With 10 values the median is the mean of the 5th and 6th: (50 + 52)/2 = 51.
Source: APPSC Assistant Statistical Officer 2025, Paper II (Notification 12/2024), Q75, final key
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On the basis of the following statements,
1. Second quartile is same as mean;
2. Second quartile is same as median;
3. Median is same as first quartile,
Which of the following holds?
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Answer C. Only 2 is true
The second quartile Q2 divides the data into two halves, so it is the median. It equals the mean only for symmetric data, and the median is not the first quartile.
Source: APPSC Assistant Statistical Officer 2025, Paper II (Notification 12/2024), Q99, final key
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Given below are two statements labelled as Assertion (A) and Reasoning (R).
Assertion (A): For a grouped frequency distribution table with one or both of the terminal classes as open, median can be computed.
Reasoning (R): Mode can be computed in case of open ended terminal classes.
In the light of the above, choose the correct option.
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Answer B. Both A and R are true and R is not correct explanation of A.
Median and mode depend only on the middle or modal class, so both can be found with open-ended terminal classes; but the fact about the mode does not explain why the median can be found.
Source: APPSC Assistant Statistical Officer 2025, Paper II (Notification 12/2024), Q114, final key
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For the following frequency distribution table, the third quartile is equal to:
Class-Intervals 60-65 65-70 70-75 75-80 80-85 Frequency 3 8 13 12 4 Show answerHide answer
Answer B. 77.5
N = 40, so Q3 is the 30th value; cumulative frequencies are 3, 11, 24, 36, so Q3 lies in 75-80. Q3 = 75 + (30 - 24)/12 x 5 = 77.5.
Source: APPSC Assistant Statistical Officer 2025, Paper II (Notification 12/2024), Q116, final key
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The weights (in grams) of 100 fruits are grouped below. Find the 65th percentile (P65).
Weight: 100–120, 120–140, 140–160, 160–180, 180–200
No. of fruits: 12, 20, 30, 26, 12
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Answer D. 162.31
65N/100 = 65; c.f.: 12, 32, 62, 88, 100, so the P65 class is 160–180. P65 = 160 + (65 − 62)/26 × 20 = 162.31 g.
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
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Find the median of the following distribution.
Class: 10–20, 20–30, 30–40, 40–50, 50–60
Frequency: 5, 7, 12, 10, 6
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Answer A. 36.67
N = 40, so N/2 = 20. Cumulative frequencies are 5, 12, 24, 34, 40, so the median class is 30–40 with L = 30, cf = 12, f = 12, h = 10. Median = L + (N/2 − cf)/f × h = 30 + (8/12) × 10 = 36.67.
Source: Standard method; answer and every option computed and checked in Python
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Find the median of the following distribution.
Class: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60
Frequency: 7, 11, 18, 24, 13, 7
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Answer A. 31.67
N = 80, N/2 = 40; c.f.: 7, 18, 36, 60, 73, 80, so the median class is 30–40. Median = L + (N/2 − c)/f × h = 30 + (40 − 36)/24 × 10 = 31.67.
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
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In the distribution below, which class contains the third quartile?
Class: 0–10, 10–20, 20–30, 30–40, 40–50
Frequency: 10, 15, 25, 20, 10
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Answer B. 30–40
N = 80, so Q3 is at 3N/4 = 60. Cumulative frequencies are 10, 25, 50, 70, 80; the first one that is at least 60 is 70, for the class 30–40.
Source: Standard method; answer and every option computed and checked in Python
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Find the fourth decile (D4) of the following distribution.
Class: 0–8, 8–16, 16–24, 24–32, 32–40, 40–48
Frequency: 9, 15, 26, 22, 18, 10
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Answer D. 20.92
N = 100, 4N/10 = 40; c.f.: 9, 24, 50, 72, 90, 100, so the D4 class is 16–24. D4 = 16 + (40 − 24)/26 × 8 = 20.92.
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
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Find the mode of the following distribution of marks.
Marks: 10–19, 20–29, 30–39, 40–49, 50–59
No. of students: 6, 15, 28, 19, 7
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Answer A. 35.41
Convert to class boundaries (subtract 0.5): modal class 29.5–39.5, f1 = 28, f0 = 15, f2 = 19, h = 10. Mode = 29.5 + 13/22 × 10 = 35.41.
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
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Which average is most appropriate for a frequency distribution with open-end classes such as 'below 10' and '60 and above'?
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Answer C. Median
Open-end classes have no definite mid-points, so the mean, GM and HM cannot be computed without assumptions. The median can be calculated as long as the median class is not open-ended.
Source: Telugu Akademi / APSCHE B.Com Business Statistics, Measures of Central Tendency
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