Question

The randomly selected delivery times (in minutes) of two fast food restaurants are as follows.
Fast food 1
42
38
37
39
38
40
45
41
40
Fast food 2
40
37
40
41
38
46
41
37
40

Answer 1

To compare the delivery times of the two fast food restaurants, we can calculate the mean and standard deviation for each set of data.

For Fast food 1:
Mean = (42 + 38 + 37 + 39 + 38 + 40 + 45 + 41 + 40) / 9
Mean = 360 / 9
Mean = 40 minutes

Standard Deviation = √[((42-40)^2 + (38-40)^2 + (37-40)^2 + (39-40)^2 + (38-40)^2 + (40-40)^2 + (45-40)^2 + (41-40)^2 + (40-40)^2) / 9]
Standard Deviation = √[(4 + 4 + 9 + 1 + 4 + 0 + 25 + 1 + 0) / 9]
Standard Deviation = √[48 / 9]
Standard Deviation = √5.33
Standard Deviation = 2.31 minutes

For Fast food 2:
Mean = (40 + 37 + 40 + 41 + 38 + 46 + 41 + 37 + 40) / 9
Mean = 360 / 9
Mean = 40 minutes

Standard Deviation = √[((40-40)^2 + (37-40)^2 + (40-40)^2 + (41-40)^2 + (38-40)^2 + (46-40)^2 + (41-40)^2 + (37-40)^2 + (40-40)^2) / 9]
Standard Deviation = √[(0 + 9 + 0 + 1 + 4 + 36 + 1 + 9 + 0) / 9]
Standard Deviation = √[60 / 9]
Standard Deviation = √6.67
Standard Deviation = 2.58 minutes

Therefore, the mean delivery time for both fast food restaurants is the same at 40 minutes, but the standard deviation for fast food 2 is slightly higher at 2.58 minutes compared to fast food 1's standard deviation of 2.31 minutes. This suggests that the delivery times for fast food 2 varies more compared to fast food 1.