One person wants to get a 95% z-confidence interval with a margin of error of at most 15 based on a population standard deviation of 60. what is the minimum sample size needed?

So the answer for this case would be n=62 rounded up to the nearest integer  

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

represent the sample mean for the sample  

population mean

represent the population standard deviation  

n represent the sample size (variable of interest)  

Confidence =95% or 0.95

The margin of error is given by this formula:  

(1)  

And on this case we have that ME =15, and we are interested in order to find the value of n, if we solve n from equation (1) we got:  

(2)  

The critical value for 95% of confidence interval is provided, , replacing into formula (2) we got:  

So the answer for this case would be n=62 rounded up to the nearest integer  

its definatley   this on your answer choice?

341

step-by-step explanation:

all you have to do is 9x31 (which equals 279) and then 20x31 (which equals 620). then you have to subtract. 620-279=341.