Question

Suppose 52% of the population has a college degree. If a random sample of size 563563 is selected, what is the probability that the proportion of persons with a college degree will differ from the population proportion by less than 5%5%? Round your answer to four decimal places.

Answer 1

Answer:

The value is  $P(| \^ p - p| < 0.05 ) = 0.9822$

Step-by-step explanation:

From the question we are told that

    The population proportion is  $p = 0.52$

     The sample size is  n  =  563      

Generally the population mean of the sampling distribution is mathematically  represented as

           $\mu_{x} = p = 0.52$

Generally the standard deviation of the sampling distribution is mathematically  evaluated as

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

=>      $\sigma = \sqrt{\frac{ 0.52 (1- 0.52 )}{563} }$

=>      $\sigma = 0.02106$

Generally the  probability that the proportion of persons with a college degree will differ from the population proportion by less than 5% is mathematically represented as

            $P(| \^ p - p| < 0.05 ) = P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 ))$

  Here  $\^ p$ is the sample proportion  of persons with a college degree.

So

 $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P(\frac{[[0.05 -0.52]]- 0.52}{0.02106} < \frac{[\^p - p] - p}{\sigma } < \frac{[[0.05 -0.52]] + 0.52}{0.02106} )$

Here  

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

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P[\frac{-0.47 - 0.52}{0.02106 } < Z < \frac{-0.47 + 0.52}{0.02106 }]$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P[ -2.37 < Z < 2.37 ]$

=>  $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P(Z < 2.37 ) - P(Z < -2.37 )$

From the z-table  the probability of  (Z <  2.37 ) and  (Z < -2.37 ) is

  $P(Z < 2.37 ) = 0.9911$

and

  $P(Z < - 2.37 ) = 0.0089$

So

=>$P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) =0.9911-0.0089$

=>$P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = 0.9822$

=> $P(| \^ p - p| < 0.05 ) = 0.9822$