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

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.