At an academically challenging high school, the average GPA of a high school senior is known to be normally distributed with a variance of 0.25. A sample of 20 seniors is taken and their average GPA is found to be 2.71. Develop a 90% confidence interval for the population mean GPA.

The 90% confidence interval for the population mean GPA is between 2.53 and 2.89

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(square root of the variance) of the population and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 2.71 - 0.18 = 2.53

The upper end of the interval is the sample mean added to M. So it is 2.71 + 0.18 = 2.89.

The 90% confidence interval for the population mean GPA is between 2.53 and 2.89

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hope this !

sinn is the total amount of students now and m were the students before, you subtract them both (n-m) and divide that by m. then, since to get a percentage from a fraction you have to multiply by 100, your answer should look like this: n-m/m*100

step-by-step explanation: