Bulb Sales over 60 Days: Mean, Median, Mode and Risk Claims
A hardware shop in Berhampur keeps a tally of how many LED bulbs it sells each day over a stretch of 60 trading days. The record is: 0 bulbs on 6 days, 1 bulb on 12 days, 2 bulbs on 18 days, 3 bulbs on 15 days and 4 bulbs on 9 days. No day had more than 4 sales. Before a festival, the owner makes two statements that his supplier wants verified. He says his 'average' sale is 3 bulbs a day, so three bulbs should always be kept in the display tray each morning. He also says that on more than half the days he sells at least 3 bulbs, which he argues justifies a bulk discount. The supplier wants the mean, median and modal daily sales worked out properly, the exact probability that the shop sells no bulb on a randomly chosen day, and a comment on how far these figures can be trusted for the festival week ahead.
1. Compute the mean, median and modal number of bulbs sold per day, and check the owner's claim of an 'average' of 3.
Read the worked answer
Total bulbs = (0×6) + (1×12) + (2×18) + (3×15) + (4×9) = 0 + 12 + 36 + 45 + 36 = 129.
Mean = 129/60 = 2.15 bulbs per day.
Median: n = 60, so take the mean of the 30th and 31st values. Cumulative frequencies are 6, 18, 36, 51, 60; both the 30th and 31st values fall in the '2 bulbs' group (cumulative 18 to 36), so median = 2.
Mode = value with the highest frequency = 2 bulbs (frequency 18) → mode = 2.
So mean 2.15, median 2, mode 2. The owner's 'average of 3' is wrong - no standard average from this data equals 3. He has probably looked at the day-count for 3 bulbs (15 days) instead of averaging the bulbs.
2. Test the owner's second claim: is the probability of selling at least 3 bulbs greater than one half?
Read the worked answer
'At least 3' means 3 or 4 bulbs.
Favourable days = 15 + 9 = 24 out of 60.
P(at least 3) = 24/60 = 2/5 = 0.4.
Since 0.4 < 0.5, the claim that this happens on more than half the days is false.
For comparison: P(less than 3) = 36/60 = 0.6, so low-sale days are actually the majority. A bulk discount justified on 'more than half the days' would be built on an error. P(exactly 3) = 15/60 = 0.25 and P(exactly 4) = 9/60 = 0.15.
3. Find the probability that the shop sells no bulb on a randomly chosen day. The owner also says he will 'never run out' if he stocks 4 bulbs. Comment on both statements.
Read the worked answer
P(0 bulbs) = 6/60 = 1/10 = 0.1, i.e. a 10% chance on a day picked at random from this record.
P(exactly 4) = 9/60 = 0.15; this is the probability a stock of 4 is fully used, not exhausted.
On 'never running out': running out means demand of 5 or more, which never occurred in the 60 days, so the empirical probability is 0/60 = 0. But a probability of 0 here only says not observed, not impossible - the shop's own record shows the daily figure rising to 4, and a festival week changes conditions (more footfall, higher demand) compared with ordinary trading days. The zero is based on a small sample and on a pattern that the festival itself may break.
Sound decision: use 0.1 as a working estimate for no-sale days and stock above 4 during the festival, treating the data as a guide to normal weeks, not as a guarantee for an exceptional one.