Question

Suppose 40% of the students in a university are baseball players. If a sample of 781 students is selected, what is the probability that the sample proportion of baseball players will be greater than 43%

Answer 1

`P(p>0.43) = 0.0436`

Explanation:

Given

`p = 40\%` -- proportion of baseball players

`n = 781` --- selected sample

Required

Determine P(p>43%)

First, calculate the z score using:

`Z = \frac{\^(p)-p}{\sqrt{(p(1-p))/(n)}}`

This gives:

`Z = \frac{48\%-40\%}{\sqrt{(40\% *(1 - 40\%))/(781)}}`

Convert % to decimal

`Z = \frac{0.43-0.40}{\sqrt{(0.40 *(1 - 0.40))/(781)}}`

`Z = \frac{0.43-0.40}{\sqrt{(0.40 *0.60)/(781)}}`

`Z = \frac{0.43-0.40}{\sqrt{(0.24)/(781)}}`

`Z = (0.03)/(√(0.00030729833))`

`Z = (0.03)/(0.01752992669)`

`Z = 1.71`

So:

`P(p>0.43) = P(Z>1.71)`

`P(p>0.43) = 1 - P(Z<1.71)`

`P(p>0.43) = 1 - 0.9564`

`P(p>0.43) = 0.0436`