1. The mean performance score on a physical fitness test for Division I student athletes is 947 with a population standard deviation of 205. Select a random sample of 64 of these students. Hint: we have a sample so use the standard error. What is the probability the mean of the sample is below 900

0.033316

Step-by-step explanation:

We use the z score formula to solve for this question.

Since we are given the number of samples in the question, our z score formula is given as:

z = (x-μ)/ S.E

where x is the raw score

μ is the sample mean

S.E is the Standard error.

x is the raw score = 900

μ is the sample mean = Population mean = 947

Standard error =

This is calculated as Population standard deviation/ √No of samples

= 205/√64.

= 205/8

= 25.625

We proceed to calculate the z score

z = (x-μ)/ S.E

z = 900 - 947/25.625

= -1.83415

Using the z score table for normal distribution,

P(x≤ z) = P(z ≤ -1.83) = P(x ≤ 900)

P(x<900) = 0.033316

Therefore, the probability the mean of the sample is below 900 is 0.033316

none of the above.

for one of the equations to be used in the process, there has to be a term in the equation since squared is .

since there is only a term in the first equation, we cannot derive a from completing the square.

solving with completing the square

8 2/3 is the answer

step-by-step explanation: just use a calculator