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kodGreya
1 month ago
8

The resting heart rates for 80 women aged 46–55 in a simple random sample are normally distributed, with a mean of 71 beats per

minute and a standard deviation of 6 beats per minute. Assuming a 90% confidence level (90% confidence level = z-score of 1.645), what is the margin of error for the population mean?
Mathematics
2 answers:
lawyer [12.5K]1 month ago
8 0

Details provided:
Confidence level = 90%
Mean = 71 beats per minute
Standard deviation = 6 beats per minute

The formula for margin of error is z * δ / √n.

Where δ represents the population standard deviation and n is the sample size; z denotes the corresponding z-value.

For a 90% confidence level, the z-value is 1.645.

Thus, the margin of error is calculated as 1.645 * (6/√80) = 1.645 * (6/8.94) = 1.645 * 0.671 = 1.104.

Svet_ta [12.7K]1 month ago
7 0

The answer:

1.10/ answer B

Detailed explanation:

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