Question

Suppose 49% of the population has a retirement account. If a random sample of size 896 is selected, what is the probability that the proportion of persons with a retirement account will differ from the population proportion by less than 3%?

Answer 1

Answer:

The probability is  $P( |\^ p - p | < 0.03 ) = 0.9275$

Step-by-step explanation:

From the question we are told that

  The population proportion is  $p = 0.49$

  The sample size is  n = 896

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

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

=>  $\sigma_{p} = \sqrt{ \frac{0.49(1 -0.49 )}{896} }$

=>  $\sigma_{p} = 0.0167$

   Generally the the probability that the proportion of persons with a retirement account will differ from the population proportion by less than 3% is mathematically represented as

    $P( |\^ p - p | < 0.03 ) = P( \frac{ | \^ p - p|}{\sigma_p } < \frac{0.03}{0.0167} )$

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

=> $P( |\^ p - p | < 0.03 ) = P(|Z| < 1.796 )$

=>  $P( |\^ p - p | < 0.03 ) = P(- 1.796 < Z < 1.796 )$

=>   $P( |\^ p - p | < 0.03 ) = P(1.796 < Z) - P(Z < -1.796 )$

From the z table  the area under the normal curve to the left corresponding to  1.796 and  - 1.796  is  

     $P(Z < -1.796 ) = 0.036247$

    $P(1.796 < Z) = 0.96375$

So

    $P( |\^ p - p | < 0.03 ) = 0.96375 - 0.036247$

=>  $P( |\^ p - p | < 0.03 ) = 0.9275$