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

Answer:

$P(p>0.43) = 0.0436$

Step-by-step explanation:

Given

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

$n = 781$ --- selected sample

Required

Determine P(p>43%)

First, calculate the z score using:

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

This gives:

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

Convert % to decimal

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

$Z = \frac{0.43-0.40}{\sqrt{\frac{0.40 *0.60}{781}}}$

$Z = \frac{0.43-0.40}{\sqrt{\frac{0.24}{781}}}$

$Z = \frac{0.03}{\sqrt{0.00030729833}}$

$Z = \frac{0.03}{0.01752992669}$

$Z = 1.71$

So:

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

$P(p>0.43) = 1 - P(Z$

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

$P(p>0.43) = 0.0436$