Question

One environmental group did a study of recycling habits in a California community. It found that 75% of the aluminum cans sold in the area were recycled. (Use the normal approximation. Round your answers to four decimal places.)
(a) If 387 cans are sold today, what is the probability that 300 or more will be recycled?



(b) Of the 387 cans sold, what is the probability that between 260 and 300 will be recycled?

Answer 1

Answer:

a

  $P( \^ p > 0.775 ) = 0.12798$

b

 $P( 0.6718 < p < 0.775 ) =0.87183$

Step-by-step explanation:

From the question we are told that

    The population proportion is  $p = 0.75$

Considering question a  

     The sample size is  $n = 387$

Generally the standard deviation of this sampling distribution is  

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

=>      $\sigma = \sqrt{ \frac{0.75(1 - 0.75)}{ 387 } }$    

=>      $\sigma = 0.022$    

The sample proportion of cans that are recycled is

                 $\^ p = \frac{ 300}{387 }$

=>              $\^ p = 0.775$

Generally the probability that 300 or more will be recycled is mathematically represented as

         $P( \^ p > 0.775 ) = P( \frac{\^ p - p }{ \sigma } > \frac{0.775 - 0.75 }{ 0.022} )$

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

       $P( \^ p > 0.775 ) = P( Z > 1.136 )$

From the z table  the area under the normal curve to the left corresponding to  1.591   is

      $P( Z > 1.136) = 0.12798$

=>    $P( \^ p > 0.775 ) = 0.12798$

Considering question b

Generally the lower limit of  sample proportion of cans that are recycled is

                 $\^ p_1 = \frac{ 260 }{387 }$

=>              $\^ p_1 = 0.6718$

Generally the upper limit of  sample proportion of cans that are recycled is

                 $\^ p_2 = \frac{ 300}{387 }$

=>              $\^ p_2 = 0.775$

Generally probability that between 260 and 300 will be recycled is mathematically represented as

           $P( 0.6718 < p < 0.775 ) = P( \frac{0.6718 - 0.75 }{ 0.022}< \frac{\^ p - p }{ \sigma }$

=>      $P( 0.6718 < p < 0.775 ) = P( -3.55 < Z < 1.136 )$

=>        $P( 0.6718 < p < 0.775 ) = P(Z < 1.136 ) - P( Z < -3.55 )$

From the z table  the area under the normal curve to the left corresponding to  1.136 and  -3.55  is

       $P( Z < -3.55 ) = 0.00019262$

and

       $P(Z < 1.136 ) = 0.87202$

So

       $P( 0.6718 < p < 0.775 ) = 0.87202- 0.00019262$

=>   $P( 0.6718 < p < 0.775 ) =0.87183$