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

Answers

Answer 1

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]


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Answers

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