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DIA [1.3K]
3 years ago
8

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
Mathematics
1 answer:
OLEGan [10]3 years ago
6 0

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

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