Answer:
The probability is [tex]P( p < 0.9207) = 0.0012556[/tex]
Step-by-step explanation:
From the question we are told
The population proportion is [tex]p = 0.96[/tex]
The sample size is [tex]n = 227[/tex]
The number of graduate who had job is k = 209
Generally given that the sample size is large enough (i.e n > 30) then the mean of this sampling distribution is
[tex]\mu_x = p = 0.96[/tex]
Generally the standard deviation of this sampling distribution is
[tex]\sigma = \sqrt{\frac{p (1 - p )}{n} }[/tex]
=> [tex]\sigma = \sqrt{\frac{0.96 (1 - 0.96 )}{227} }[/tex]
=> [tex]\sigma = 0.0130[/tex]
Generally the sample proportion is mathematically represented as
[tex]\^ p = \frac{k}{n}[/tex]
=> [tex]\^ p = \frac{209}{227}[/tex]
=> [tex]\^ p = 0.9207[/tex]
Generally probability of obtaining a sample proportion as low as or lower than this, if the university’s claim is true, is mathematically represented as
[tex]P( p < 0.9207) = P( \frac{\^ p - p }{\sigma } < \frac{0.9207 - 0.96}{0.0130 } )[/tex]
[tex]\frac{\^ p - p}{\sigma } = Z (The \ standardized \ value\ of \ \^ p )[/tex]
[tex]P( p < 0.9207) = P(Z< -3.022 )[/tex]
From the z table the area under the normal curve to the left corresponding to -3.022 is
[tex]P(Z< -3.022 ) = 0.0012556[/tex]
=> [tex]P( p < 0.9207) = 0.0012556[/tex]
