Question

If the amount of change a statistics student carries has a mean of $0.88 and a standard deviation of $0.30, suppose that we randomly pick 36 online statistics students. Find the probability the average of 36 students is between $.80 and $1. The Probability that an individual student has between $.80 and $1 is .9370.

Answer 1

We have the following:

We must calculate the value of z

a.

`Z=(x-m)/(s)`

x, is the value to evalute (0.8 - 1)

m, is the mean (0.88)

s, is the standard deviation (0.3)

n, is the sample size (36)

Therefore, the probability is 0.258 or 25.8%

b.

`Z=\frac{x-m}{\frac{s}{\sqrt[]{n}}}`

x, is the value to evalute (0.8 - 1)

m, is the mean (0.88)

s, is the standard deviation (0.3)

n, is the sample size (36)

replacing:

Therefore, the probability is 0.937 or 93.7%