Question

Ms. Miller, the manager of a store, wants to find out what percent of all customers spend more than $50$50 on a single visit.



She randomly collects 3535 receipts from last year.

The rounded amounts, in dollars, are shown below.





10,45,56,63,72 25,46,56,64,74 29,49,57,65,74 37,52,57,65,76 39,53,58,66,79 41,53,61,66 ,81 43,55,63,67,95





Part A:

Find the proportion of customers who spend more than $50$50 on a single visit.






Part B:

Calculate the mean and standard deviation of the amount of money spent in a single visit (rounded to the nearest cent).

Mean ≈ $


Standard deviation ≈ $




Part C:

Identify the range of amounts of money spent in a single visit within 22 standard deviations of the sample mean.



Lower range = $
,

Upper range = $

Answer 1

Part A:
Number > $50: 25 out of 35.
Proportion = 25/35 = 5/7 ≈ 0.7143 (≈ 71.43%).

Part B:
Mean = 1992/35 ≈ $56.91.
Sample standard deviation ≈ $16.96.

Part C:
Within 2 standard deviations: mean ± 2s
Lower = 56.91 − 2(16.96) ≈ $22.99
Upper = 56.91 + 2(16.96) ≈ $90.84