Assistant Provident Fund Commissioner and Enforcement Officer / Accounts Officer (APFC and EO/AO) 2025 recruitment test. Transcribed from the Commission's own booklet (series A, PENA-A-PTS); answer worked out by our faculty — UPSC has published no answer key for this sitting. This question is in Part B. Answer last reviewed 25 August 2026.

The scores of 16 students in a test are as follows: 19, 4, 17, 7, 15, 2, 18, 11, 15, 17, 19, 4, 3, 2, 6, 9. What is the difference between their arithmetic mean and the median ?

(a)1·0
(b)0·8
(c)0·5
(d)0·2

Answer

Why

Correct — C, (c) 0·5. The mean is 10·5, the median is 10, and the difference is 0·5.

THE ARITHMETIC MEAN. Add the sixteen scores: 19 + 4 = 23; + 17 = 40; + 7 = 47; + 15 = 62; + 2 = 64; + 18 = 82; + 11 = 93; + 15 = 108; + 17 = 125; + 19 = 144; + 4 = 148; + 3 = 151; + 2 = 153; + 6 = 159; + 9 = 168. The total is 168, and there are 16 scores, so the mean is 168 ÷ 16 = 10·5. Adding a long list in one pass invites error, so it is worth pairing terms that make round numbers — 19 + 11, 17 + 3, 18 + 2, 4 + 6 and so on — and keeping a count of how many values have been used, which also guards against dropping one.

THE MEDIAN. Arrange the scores in order: 2, 2, 3, 4, 4, 6, 7, 9, 11, 15, 15, 17, 17, 18, 19, 19. Count them: sixteen, so nothing has been lost in sorting. With an EVEN number of observations the median is the mean of the two middle ones, that is of the 8th and the 9th values. The 8th is 9 and the 9th is 11, so the median is (9 + 11) ÷ 2 = 10.

THE DIFFERENCE. 10·5 − 10 = 0·5.

Two cautions about the data. The list contains REPEATED values — 19, 17, 15, 4 and 2 each appear twice — and every repetition must be carried through into the ordered list; deleting duplicates would leave eleven values and a different median altogether. And with sixteen values, the median is not a value in the list at all but the average of two of them; taking the 8th value alone would give 9, and taking the 9th alone would give 11, neither of which is the median.

Why the others are wrong

(a)1·0 — A difference of one arises from taking the median as 9·5 or the mean as 11 — that is, from a slip of one place in the ordered list, or from an error of eight in the total. It is also what a candidate gets by treating the median as the 8th value in a list of sixteen rather than as the average of the 8th and 9th, and then compensating elsewhere. Recount the ordered list to sixteen before locating the middle.
(b)0·8 — Requires a mean of 10·8, that is a total of 172·8 — not a whole number, so it cannot come from sixteen whole-number scores at all. This option is there to be chosen by a candidate who has approximated the division rather than carrying it out; 168 ÷ 16 is exactly 10·5, and any mean of sixteen integers must be a whole number of sixteenths.
(d)0·2 — Would need a mean of 10·2, that is a total of 163·2 — again impossible for sixteen whole-number scores. The value typically appears when a candidate drops one of the repeated scores in copying the list and then divides by the original count of sixteen, producing a mean slightly below the true one. Copy the list once, count the entries, and only then compute.

Concept

The mean, the median and the mode are the three measures of central tendency and each answers a slightly different question. The MEAN is the total divided by the number of observations; it uses every value and is therefore sensitive to extremes. The MEDIAN is the middle value of the ordered data — the (n+1)/2th observation when n is odd, and the average of the n/2th and (n/2 + 1)th when n is even; it is unaffected by how extreme the extremes are. The MODE is the most frequently occurring value. In a symmetric distribution all three coincide; where the data are skewed they separate, and the direction of the gap between mean and median indicates which way. Here the mean sits slightly above the median, reflecting a mild pull from the cluster of high scores.

Statistics items in these papers are always computational and always short, and the risk in them is clerical rather than conceptual: a mis-added total or a mis-sorted list. Two disciplines defeat that risk — count the observations after sorting, and check that the answer is of a possible form. Two of the wrong options here imply a mean that sixteen whole-number scores could not produce at all, which is a check worth making before any arithmetic is repeated.

Statistics items on this paper are short and computational, and they are lost to clerical error rather than to ignorance. Two defences are worth building into the routine. First, count the observations after sorting them and check the count against the stem, which catches a value dropped in copying — the risk here is real, since five of the sixteen scores are repeated. Second, ask whether the answer is of a possible form: a mean of sixteen whole numbers must be a whole number of sixteenths, which quietly eliminates two of the four options before any arithmetic is checked.

Key facts

The sixteen scores total 168, so the mean is 168 ÷ 16 = 10·5.
Sorted, the scores are 2, 2, 3, 4, 4, 6, 7, 9, 11, 15, 15, 17, 17, 18, 19, 19.
For an even number of observations the median is the average of the two middle values.
The 8th and 9th values are 9 and 11, so the median is 10.
The difference between mean and median is 0·5.
Repeated values must be retained in the ordered list; the data contain five repeated scores.
The mean uses every observation and is affected by extreme values; the median is not.
In this data set the mean exceeds the median, indicating a slight pull from the higher scores.

Study next

Mean, median and mode — definitions, computation and when each is preferredMedian for grouped data and the cumulative frequency methodRange, quartiles and standard deviation as measures of dispersionWeighted mean and combined mean of two groupsThe effect of an outlier on each measure of central tendency

Common traps

Taking the median as a single middle value when the number of observations is even.
Dropping repeated values when sorting the data.
Mis-adding a long inline list; pair terms into round numbers and keep a running count.
Approximating the division instead of carrying it out exactly.

Statistics items in EPFO papers give a short data set and ask for one measure, or for the difference between two. They are quick marks and are lost almost entirely to clerical slips, so the sorting and the addition deserve more care than the concepts do. Expect a single data set with a single question — mean, median, or the gap between them — and treat the sorting and addition as the substance of the item, since that is where the marks are actually decided.

Related PYQs

EPFO_EOAO_2020_Q119

The average weight of 100 students in a class is 46 kg. The average weights of boys and girls are 50 kg and 40 kg respectively. What is the difference between the number of boys and girls ?

(a) 30
(b) 25
(c) 20
(d) 10

Answer(c) 20

An averages item from the earlier paper built on the same definition — the average weight of a class and of its two groups, used to find the difference between the numbers of boys and girls.

Practice

practice — not a real PYQ
What is the median of the observations 4, 9, 2, 7, 5, 6 ?
- (a)5
- (b)5·5
- (c)6
- (d)6·5
Answer(b) 5·5
practice — not a real PYQ
The mean of five observations is 12. If four of them are 8, 10, 14 and 16, what is the fifth ?
- (a)10
- (b)12
- (c)14
- (d)16
Answer(b) 12

Why