Ordering loaves from sixteen days of canteen sales
The canteen at a Windhoek secondary school recorded the number of loaves of bread sold on each of 16 trading days: 42, 45, 44, 43, 46, 45, 44, 47, 45, 43, 44, 46, 45, 44, 45, 120.
The 120 was the day of the inter-house athletics meeting, when visiting teams bought in bulk. The manager must decide how many loaves to order per day next term and has been told by the supplier to base the order on 'the average'. One learner uses the mean of all 16 days; another says the athletics day must be excluded; a third wants a probability statement about demand on two randomly chosen days. Bread not sold on the day is given to the hostel at no income, so ordering too high has a real cost, while running out loses sales.
1. Calculate the mean and the median of all 16 days and decide which better represents a typical trading day.
Read the worked answer
Sum of all 16 values = 788, so mean = 788 ÷ 16 = 49.25 loaves.
Sorted data: 42, 43, 43, 44, 44, 44, 44, 45, 45, 45, 45, 45, 46, 46, 47, 120. The 8th and 9th values are both 45, so median = 45 loaves. The mode is also 45, occurring on 5 days.
The 120 day sits 120 − 45 = 75 above the typical value; spread over 16 days that adds 75 ÷ 16 ≈ 4.7 loaves to every day's mean, which is exactly the gap between 49.25 and 45.
Judgement: the median (and the mode) describe a normal trading day; the mean is pulled up by one bulk day. Ordering 49 loaves daily would leave about 4 unsold loaves on a normal day, which are given away, so the median is the better basis - and 45 also matches the most frequent outcome.
2. Recalculate after excluding the athletics day and explain whether excluding it is a defensible decision for stock planning.
Read the worked answer
Excluding the 120: sum = 668 over 15 days, mean = 668 ÷ 15 = 44.53; the median of the 15 values is the 8th value = 45.
Mean and median are now close, which confirms that the single 120 day caused the earlier disagreement.
Is exclusion defensible? Only partly. The 120 is not a recording error or a misprint - bulk demand from visiting teams is real demand and will recur at future events. Deleting it is legitimate for estimating an ordinary day's sales, but not for saying the canteen never needs more than 45 loaves.
Practical decision: order the typical 45 loaves daily and add an extra batch on known event days. If the manager orders exactly 45 every day, sales exceeded 45 on 4 of the 16 days (46, 47, 46 and 120), so she will run short on those days. Choosing 45 is therefore a decision about acceptable risk, not a calculation that removes the risk.
3. Two of the 16 days are chosen at random without replacement. Find the probability that bread sales were at least 45 on both days.
Read the worked answer
Count the days with sales ≥ 45: 45, 46, 45, 47, 45, 46, 45, 45 and 120 - that is 9 of the 16 days.
First day: P(≥45) = 9/16.
Second day, without replacement: if the first day qualified, only 8 of the remaining 15 days qualify, so P(both) = 9/16 × 8/15 = 72/240 = 0.30.
Check: the multiplication rule applies because the two selections are not independent - the first draw changes the composition of the remaining days.
Judgement: using 9/16 × 9/16 = 0.317 would double-count the same day twice and is wrong for sampling without replacement. Also note the limits of the estimate: with only 16 observations, 0.30 is an estimate, and a longer record could reasonably give 0.25 or 0.35. The probability describes the past pattern, not a guarantee about next term's event days.