Question

# 5 in a local university, 40% ofthe students live on the campus. A random sample of 96 tudents is selected for a special study. 쩍-96 41 ates. n error ofproportion-....6千 A. What is the probability that the sample proportion living at campus is between 0. 30 and 0. 50?

Answer 1

Standard error is

$\sqrt{\frac{p*(1-p)}{n}}$

$\sqrt{\frac{0.4*(1-0.4)}{96}}=0.05$

A) We have to find P( 0.30 < $\hat{p}$ < 0.50)

For finding this probability we have to find z score.

$z=\frac{\hat{p}-p}{SE}=\frac{0.30-0.40}{0.05}=-2$

$z=\frac{\hat{p}-p}{SE}=\frac{0.50-0.40}{0.05}=2$

That is we have to find P( - 2 < Z < 2)

P( - 2 < Z < 2) = P(Z < 2) - P(Z < -2) = 0.9772 - 0.0225 = 0.9545