The pulse rate (in bpm) of a random sample of 30 Peruvian Indians was collected. The mean pulse rate of the sample is 70.2, with a sample standard deviation of 10.51. Compute a 95% confidence interval for the population mean. Assume the pulse rate is normally distributed

The 95% confidence interval for the population mean is between 66.44 bpm and 73.96 bpm.

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find M as such

In which is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the mean subtracted by M. So it is 70.2 - 3.76 = 66.44 bpm

The upper end of the interval is the mean added to M. So it is 70.2 + 3.76 = 73.96 bpm

The 95% confidence interval for the population mean is between 66.44 bpm and 73.96 bpm.

answer: 193

explanation:

200-7 = 193