A Stationery Shop's Smudged Register: Missing Frequency, Median versus Mean
The stationery shop outside a school records the number of notebooks sold each day for 30 days, grouped into classes 0–10, 10–20, 20–30, 30–40 and 40–50. The owner has written the mean as 23 notebooks per day, but the entry for the 20–30 class has been smudged and cannot be read. The other entries are 4, 8, ?, 6 and 2 respectively. The owner wants to print a sign saying "about 23 notebooks sold daily" to attract a wholesaler, while a supplier argues that two days of bulk orders, both falling in the 40–50 class, have inflated the figure and that the true typical day is much lower. The shop's accountant says the median should be quoted instead of the mean whenever there are unusual days, but the owner insists the two bulk days are genuine business and must count. The shop has no record of the exact value of any single day's sales, only the class totals.
1. Use the stated mean to recover the missing frequency, and verify the mean.
Read the worked answer
Let the missing frequency be f. Midpoints are 5, 15, 25, 35, 45.
Σfx = 4(5) + 8(15) + f(25) + 6(35) + 2(45) = 20 + 120 + 25f + 210 + 90 = 440 + 25f. Total frequency = 4 + 8 + f + 6 + 2 = 20 + f.
Mean = 23 gives 440 + 25f = 23(20 + f) = 460 + 23f, so 2f = 20 and f = 10.
Check: total days = 30 (matches the register), Σfx = 440 + 250 = 690, mean = 690/30 = 23. Consistent.
The judgement point is that the mean fixes only one unknown; with two unknowns the register would be under-determined and the owner's figure could not be trusted at all.
2. Compute the median and the mode, and say whether they support the owner's sign.
Read the worked answer
Cumulative frequencies: 4, 12, 22, 28, 30. Here n = 30, so n/2 = 15, which first falls in the class 20–30 (cf 22 > 15, previous cf 12).
Median = L + [(n/2 − cf)/f] × h = 20 + [(15 − 12)/10] × 10 = 20 + 3 = 23 notebooks.
Mode: modal class 20–30 (frequency 10). Mode = L + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h = 20 + [(10 − 8)/(20 − 8 − 6)] × 10 = 20 + (2/6) × 10 ≈ 23.33 notebooks.
All three averages agree to within a third of a notebook, so "about 23 notebooks sold daily" is a fair description of the grouped data. The sign is defensible, but the owner should print it as an average of grouped figures, since no exact daily value is known.
3. Test the supplier's claim that the bulk days inflate the mean badly. Estimate the mean with those two days removed, and compare how the mean and median respond.
Read the worked answer
The two bulk days lie in 40–50, whose class midpoint is 45, so their combined contribution is about 2 × 45 = 90 notebooks.
Remove them: Σfx ≈ 690 − 90 = 600 over 30 − 2 = 28 days, giving mean ≈ 600/28 ≈ 21.4 notebooks.
The mean falls by about 1.6 notebooks, a drop of roughly 7%. That is real but modest: it does not support the supplier's description of a badly inflated figure.
Response of the two averages: the mean uses every value and every magnitude, so two extreme days shift it; the median depends on position only. Removing the two highest days leaves 28 values, and the median of the reduced data stays in the 20–30 class, near 22–23, barely moved.
Judgement: both figures describe the same shop, and neither is wrong. The owner is right that genuine bulk sales belong in a total or a mean used for stocking decisions; the accountant is right that a median is the safer "typical day" figure when a wholesaler will read the sign as a promise about ordinary trading. The honest sign would quote both.