Bunuel
A supermarket chain randomly selected 40 customer purchases from each of 10 stores. For each purchase, the number of items bought was recorded. The table shows, for each store, how many of the 40 purchases contained a number of items within each indicated range.
| Number of items purchased |
| Store | 1 to 4 | 5 to 8 | 9 to 12 | 13 to 16 | 17 to 20 | 21 to 24 |
|---|
| A | 12 | 10 | 8 | 5 | 3 | 2 |
| B | 15 | 8 | 7 | 5 | 4 | 1 |
| C | 9 | 12 | 10 | 5 | 3 | 1 |
| D | 18 | 7 | 5 | 4 | 4 | 2 |
| E | 3 | 4 | 5 | 8 | 9 | 11 |
| F | 11 | 11 | 9 | 5 | 3 | 1 |
| G | 14 | 9 | 8 | 4 | 3 | 2 |
| H | 0 | 0 | 12 | 12 | 10 | 6 |
| I | 6 | 9 | 11 | 7 | 5 | 2 |
| J | 8 | 8 | 9 | 7 | 5 | 3 |
For each of the following questions, select
Can be determined if the answer can be determined from the information provided. Otherwise, select
Cannot be determined.
Official Solution:The table summarizes 40 purchases at each store. Each entry tells us how many purchases contained a number of items within the corresponding range. For example, at Store D, 18 purchases contained 1 to 4 items, 7 purchases contained 5 to 8 items, and 5 purchases contained 9 to 12 items. The exact number of items in each individual purchase is not given.
• Which store had the least range in the number of items purchased?At Store H, there were no purchases in either the 1-to-4 or the 5-to-8 range. Thus, every purchase at Store H contained at least 9 items and at most 24 items.
Therefore, the range at Store H can be no greater than:
24 - 9 = 15
At every other store, there is at least one purchase in the 1-to-4 range and at least one purchase in the 21-to-24 range. Thus, at each of those stores, the range must be at least:
21 - 4 = 17
Since Store H's range is at most 15, while every other store's range is at least 17, Store H must have had the least range.
Answer:
Can be determined.
• Was the mean number of items purchased at Store E greater than 16?Because the table gives ranges rather than exact numbers, consider the smallest and largest possible totals for Store E.
Using the smallest possible number of items in each range:
3 * 1 + 4 * 5 + 5 * 9 + 8 * 13 + 9 * 17 + 11 * 21 = 556
The corresponding mean is:
556/40 = 13.9
Using the largest possible number of items in each range:
3 * 4 + 4 * 8 + 5 * 12 + 8 * 16 + 9 * 20 + 11 * 24 = 676
The corresponding mean is:
676/40 = 16.9
Thus, depending on the exact numbers of items within the ranges, the mean could be less than 16 or greater than 16.
Answer:
Cannot be determined.
• Was the median number of items purchased at most 16 at every store?Each store has 40 purchases, so the median is the average of the 20th and 21st values when the purchases are arranged from smallest to largest.
For every store except Store E, at least 21 purchases contained 16 items or fewer. Therefore, at each of those stores, both the 20th and 21st values are at most 16, so the median must be at most 16.
At Store E:
3 + 4 + 5 + 8 = 20
Exactly 20 purchases contained 16 items or fewer. Thus, the 20th value is between 13 and 16, while the 21st value is between 17 and 20.
For example, if all 8 values in the 13-to-16 range were 13, making the 20th value 13, and the 21st value were 17, the median would be:
(13 + 17)/2 = 15
which is less than 16 and therefore falls within the “at most 16” range.
However, if all 8 values in the 13-to-16 range were 16, making the 20th value 16, and the 21st value were 20, the median would be:
(16 + 20)/2 = 18
which is greater than 16.
Therefore, because the median at Store E could be at most 16 or greater than 16, we cannot determine whether the median was at most 16 at
every store.
Answer:
Cannot be determined.