Question

A selective university advertises that 96% of its bachelor’s degree graduates have, on graduation day, a professional job offer or acceptance in a graduate degree program in their major area of study. In a sample of 227 recent graduates this was true of 209 of them. The probability of obtaining a sample proportion as low as or lower than this, if the university’s claim is true, is about:________
a. 0.015
b. 0.001
c. 0.131
d, 0.084

Answer 1

Answer:

The probability is  $P( p < 0.9207) = 0.0012556$

Step-by-step explanation:

From the question we are told

  The population proportion is $p = 0.96$

 The sample size is  $n = 227$

 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  

       $\mu_x = p = 0.96$

Generally the standard deviation of this sampling distribution is  

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

=>  $\sigma = \sqrt{\frac{0.96 (1 - 0.96 )}{227} }$

=>  $\sigma = 0.0130$

Generally the sample proportion is mathematically represented as

      $\^ p = \frac{k}{n}$

=> $\^ p = \frac{209}{227}$

=> $\^ p = 0.9207$

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

     $P( p < 0.9207) = P( \frac{\^ p - p }{\sigma } < \frac{0.9207 - 0.96}{0.0130 } )$

$\frac{\^ p - p}{\sigma } = Z (The \ standardized \ value\ of \ \^ p )$

   $P( p < 0.9207) = P(Z< -3.022 )$

From the z table  the area under the normal curve to the left corresponding to    -3.022  is

     $P(Z< -3.022 ) = 0.0012556$

=> $P( p < 0.9207) = 0.0012556$