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Employed Graduates: According to the administration at a particular college, 55% of graduates are typically employed full time after graduation. To assess the impact of the current job market, the administration will take a survey of 619 recent graduates. The administration finds that 52% of the surveyed recent graduates were employed full time. If the population proportion of all recent graduates employed is the same as the stated proportion typically employed, how likely is it that the administration would have obtained a sample proportion as low as or lower than it obtained from its survey

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okpalawalter8

Answer:

The likelihood is    [tex]P(p <  \^ p) = 0.066807[/tex]

Step-by-step explanation:

From the question we are told that  

   The population proportion is p =0.55

    The sample size is  n  =  619  

   The sample proportion is  [tex]\^ p =  0.52[/tex]

Generally the mean of the sampling distribution is  [tex]\mu_{x} =  p = 0.55[/tex]

The standard deviation is  [tex]\sigma =  \sqrt{\frac{p(1- p )}{n} }[/tex]

=>    [tex]\sigma =  \sqrt{\frac{0.55(1- 0.55 )}{619} }[/tex]

=>     [tex]\sigma =  0.02[/tex]

Generally the likelihood that  the administration have obtained a sample proportion as low as or lower than it obtained from its survey is mathematically represented as

[tex]P(p <  \^ p) =  P(\frac{p- \mu}{\sigma }  < \frac{ 0.52- 0.55}{0.02 }   )[/tex]

Generally [tex]\frac{p- \mu}{\sigma }  =  Z(The \  standardized \  value  \  of  \  p)[/tex]

=>   [tex]P(p <  \^ p) =  P(Z <-1.5  )[/tex]

From the z-table the probability of (Z <-1.5  ) is  

       [tex]P(Z <-1.5  )  = 0.066807[/tex]

So

      [tex]P(p <  \^ p) = 0.066807[/tex]

 

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