Question

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?

Answer 1

Hey mate,

◆ Answer- 1.1035 bpm

● Explaination-
# Given-
z = 1.645
σ = 6 bpm
μ = 71 bpm
n = 80 people

# Solution-
Margin of error is given by-
Margin of error = zσ/√n
Margin of error = 6×1.645/√80
Margin of error = 6×1.645/√80
Margin of error = 1.1035 bpm

Thus margin of error if given sample with 90% C.I. is 1.1035 bpm.

Hope this is helpful...

Answer 2

1.1035

just took the test