Homework SourseA random sample of 25 customers was chosen in a MiniMart bet
A random sample of 25 customers was chosen in a MiniMart between 3:00 and 4:00pm on a Friday afternoon. The frequency distribution below shows the distribution for checkout time (in minutes).
Total 25
a. Complete the frequency table with frequency and relative frequency.
b. What percentage of the checkout times was less than 3 minutes?
c. In what class interval must the median lie? Explain your answer
d. Assume that the largest observation in this dataset is 5.8. Suppose this observation were incorrectly recorded as 8.5 instead of 5.8, will the mean increase, decrease or remain the same? Will the median increase, decrease or remain the same?
| Checkout Time (in minutes) | Frequency | Relative Frequency |
| 1.0-1.9 | 2 | |
| 2.0-2.9 | 8 | |
| 3.0-3.9 | ||
| 4.0-5.9 | 5 |
Solution
b) <3 minutes = 10/25 = 2/5
c) Median lies in the middle.
If we calculate cumulative frequency, we see that 12.5 (half of 25) lie in the interval 3.0 to 3.9
d) If 5.8 is recorded wrongly as 8.5. sum of all entries will decrease by 2.7
Hence mean will decrease , but median will not change as the highest entry change willnot affect median.
| Checkout Time (in minutes) | Frequency | Relative Frequency |
| 1.0-1.9 | 2 | 2/25 |
| 2.0-2.9 | 8 | 8/25 |
| 3.0-3.9 | 10 | 2/5 |
| 4.0-5.9 | 5 | 1/5 |
| Total 25 |
