Question

Suppose 41% of the students in a university are baseball players. If a sample of 524 students is selected, what is the probability that the sample proportion of baseball players will be greater than 44%

Answer 1

"0.0808" is the appropriate response.

Explanation:

Given:

n = 524

`\hat{P}` = 41%

or,

= 0.41

`1-\hat{P}=1-0.41`

`=0.59`

`\mu \hat{P}=\hat{P}`

`=0.41`

Now,

⇒
`6 \hat{P}=\sqrt{\frac{\hat {P}(1-\hat{P})}{n} }`

`=\sqrt{(0.41* 0.59)/(524) }`

`=0.0215`

`P(\hat {P}>44 \ percent)`

or,

`P(\hat{P}>0.44)`

`=1-P(\hat{P}<0.44)`

`=1-P(\frac{\hat{P}-\mu \hat{P}}{6 \hat{P}} <(0.44-0.41)/(0.0215) )`

`=1-P(z<1.40)`

By using the standard normal table, we get

`=1-0.9192`

`=0.0808`