A random sample of 25 customers was chosen in CCP MiniMart between 3:00 and 4:00 PM on a Friday afternoon. The frequency distribution below shows the distribution for checkout time (in minutes).
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 | ?
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? Why?
(a)
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 (since 25 - 5 + 8 + 2) = 10 | 10/25
4.0 - 5.9 | 5 | 5/25
Total | 25 | ?
(b) (2 + 8)/25 = 10/25 = 40%
c) 3.0 - 3.9 since 2 + 8 + 10 + 5 = 25 and 13 is the middle value which occurs in the 3.0 - 3.9 interval
(d) Mean increases since it's a higher value than usual. Median would not change as the median is the most frequent distribution and assuming the 5.8 is only recorded once.
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 | ?
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? Why?
(a)
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 (since 25 - 5 + 8 + 2) = 10 | 10/25
4.0 - 5.9 | 5 | 5/25
Total | 25 | ?
(b) (2 + 8)/25 = 10/25 = 40%
c) 3.0 - 3.9 since 2 + 8 + 10 + 5 = 25 and 13 is the middle value which occurs in the 3.0 - 3.9 interval
(d) Mean increases since it's a higher value than usual. Median would not change as the median is the most frequent distribution and assuming the 5.8 is only recorded once.