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, what is the margin of error for the population mean? 0.66 1.10 1.31 1.73

Given:
Confidence level = 90%
mean = 71 beats per minute
standard deviation = 6 beats per minute

margin of error = z * δ / √n

where : δ - population of the standard deviation, n is the sample size ; z is the appropriate z value.

90% confidence level = 1.645 in z-value

margin of error = 1.645 * (6/√80) = 1.645 * (6/8.94) = 1.645 *  0.671 = 1.104

The margin of error is 1.10