Question

Lunch break: In a recent survey of 611 working Americans ages 25–34, the average weekly amount spent on lunch as $45.62 with standard deviation $2.88. The weekly amounts are approximately bell–shaped. Part 1 Estimate the percentage of amounts that were less than $39.86. Round the answer to one decimal place. Approximately 2.3 % of the amounts were less than $39.86. Correct answer: 2.5 Part 2 out of 3 Estimate the percentage of amounts that were greater than $48.50. Round the answer to one decimal place. Approximately % of the amounts were greater than $48.50.

          Lunch break: In a recent survey of 611 working Americans ages 25–34, the average weekly amount spent on lunch as $45.62 with standard deviation $2.88. The weekly amounts are approximately bell–shaped.

Part 1
Estimate the percentage of amounts that were less than $39.86. Round the answer to one decimal place.

Approximately 2.3 % of the amounts were less than $39.86.
Correct answer:
2.5

Part 2 out of 3
Estimate the percentage of amounts that were greater than $48.50. Round the answer to one decimal place.

Approximately % of the amounts were greater than $48.50.

Show more…

Lunch break: In a recent survey of 611 working Americans ages 25–34, the average weekly amount spent on lunch as 45.62 with standard deviation2.88. The weekly amounts are approximately bell–shaped.

Part 1
Estimate the percentage of amounts that were less than 39.86. Round the answer to one decimal place.

Approximately 2.3 % of the amounts were less than $39.86.
Correct answer:
2.5

Part 2 out of 3
Estimate the percentage of amounts that were greater than48.50. Round the answer to one decimal place.

Approximately % of the amounts were greater than $48.50.

Added by Thomas G.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Madhur L

07/06/2023

Step 1

The z-score is calculated as: z = (x - μ) / σ where x is the value, μ is the mean, and σ is the standard deviation. For $39.86: z = (39.86 - 45.62) / 2.88 z ≈ -1.99 For $48.50: z = (48.50 - 45.62) / 2.88 z ≈ 1.00 Now, we can use the z-score to find the Show more…

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Lunch break: In a recent survey of 611 working Americans ages 25-34, the average weekly amount spent on lunch as $45.62 with standard deviation $2.88. The weekly amounts are approximately bell-shaped. Part 1 Estimate the percentage of amounts that were less than $39.86. Round the answer to one decimal place. Approximately 2.3 % of the amounts were less than $39.86. Correct answer: 2.5 Part 2 out of 3 Estimate the percentage of amounts that were greater than $48.50. Round the answer to one decimal place. Approximately % of the amounts were greater than $48.50.

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