Raffle Sales: Which Average to Report, and a Parent's 'Certain' Win
For its annual function, a school in Prayagraj prints raffle coupons and sells them over eleven days. Daily sales were 120, 145, 135, 160, 145, 150, 145, 175, 130, 90 and 155 coupons. While reconciling the accounts, the clerk finds that the figure 90 is an entry error: on that day 190 coupons were actually sold. The school printed 2,500 coupons in all. Of these, 100 win a pen, 20 win a book and 5 win a bicycle; the rest win nothing. A parent buys 20 coupons and tells the counter staff that he is now certain to win something, because '125 coupons win and I have 20 of the 2,500'. The principal wants a short statistical note on the sales figures - mean, median and mode - and a correct statement of the parent's chance, before the draw is announced on stage.
1. Compute the mean, median and mode of the daily sales as they stand in the register (with the 90).
Read the worked answer
Sum = 120 + 145 + 135 + 160 + 145 + 150 + 145 + 175 + 130 + 90 + 155. Group it: 120 + 145 = 265; +135 = 400; +160 = 560; +145 = 705; +150 = 855; +145 = 1000; +175 = 1175; +130 = 1305; +90 = 1395; +155 = 1550.
Mean = 1550 / 11 = 140.9 (≈ 141 coupons per day).
Median: arrange in order - 90, 120, 130, 135, 145, 145, 145, 150, 155, 160, 175. There are 11 values, so the median is the 6th: 145.
Mode: 145 occurs three times, more than any other value, so the mode is 145.
Judgement: here the mean (140.9) sits below the median and mode (both 145) because the single low day of 90 drags it down. When the mean and median disagree, the median is usually the better description of a typical day, and reporting only 'about 141' would understate a normal day's sale.
2. Recalculate after the 90 is corrected to 190. Which of the three averages moves, and what does that show about choosing a summary?
Read the worked answer
Correction adds 190 − 90 = 100 to the total. New sum = 1550 + 100 = 1650; n is still 11. New mean = 1650 / 11 = 150 exactly. The mean moves by 100/11 ≈ 9.1 coupons per day.
New ordered list: 120, 130, 135, 145, 145, 145, 150, 155, 160, 175, 190. The 6th value is still 145, so the median does not change. The frequency of 145 is untouched, so the mode stays 145.
What this shows: the mean uses the size of every observation, so one mistaken entry of 100 coupons shifts it by 9.1; the median depends only on position, so pushing the largest value further out changes nothing; the mode depends only on how often a value repeats.
Judgement for the principal's note: report the median and mode (145) as the typical daily sale, and mention the mean (150) as the total-based figure. Also state that the correction must be entered in the register before any total is quoted to donors, since the eleven-day total changes from 1,550 to 1,650 coupons.
3. Correct the parent's reasoning and give the actual probability. Is 20 coupons out of 2,500 a guarantee?
Read the worked answer
Winning coupons = 100 (pen) + 20 (book) + 5 (bicycle) = 125 out of 2,500.
P(a randomly drawn coupon wins) = 125/2500 = 1/20 = 0.05. P(it wins nothing) = 1 − 0.05 = 0.95.
Where the parent's reasoning fails: he has computed 20 × (1/20) = 1 and read that as certainty. But that product is an expected number of wins over many repetitions of the whole purchase - it is not a probability, and an expectation of 1 does not mean a guaranteed 1. Each of his 20 coupons is one of 2,500 and could equally be a blank.
A quick approximate check, treating his 20 coupons as near-independent draws: P(no win at all) ≈ 0.95²⁰ ≈ 0.36, so P(at least one win) ≈ 0.64 - roughly a 36% chance of winning nothing. 'Certain' is plainly wrong.
Judgement: the honest statement for the stage announcement is that every coupon has a 1-in-20 chance and that a buyer of 20 coupons has roughly a two-in-three chance of some prize, most likely the smallest one, since only 5 of the 125 winning coupons carry the bicycle.