Question

Research in March 2019 suggests that 40% of U.S. Adults approve of way President Trump is running the country. We randomly sample 50 U.S. Adults and find that 35% approve of way President Trump is running the country. What is the probability that a random sample of 50 U.S. Adults has less than 35% with this opinion? (Round standard error to 2 decimal places before calculating Z. Round Z to 2 decimal places before using the Normal Distribution Calculator.) Group of answer choices about 0.24 about 0.76 about 0.48

Answer 1

Answer:

The probability is  $P( \^ p  <  0.35)=    0.2388$

Step-by-step explanation:

From the question we are told that

   The  population proportion is  p =  0.40

    The sample size is  n  =  50  

   

 Generally the standard error is mathematically represented as

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

=>     $\sigma_{\= p} =  \sqrt{\frac{0.40 *  (1- 0.40)}{50} }$

=>     $\sigma_{\= p} = 0.07$

Generally the probability that a random sample of 50 U.S. Adults has less than 35% with this opinion is mathematically represented as

      $P( \^ p  <  0.35) =  P(\frac{\^ p  -  p}{\sigma_{\= p}} <  \frac{0.35 - 0.40}{0.07}  )$

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

=>   $P( \^ p  <  0.35)=P(Z  <  -0.71)$

From the z table  

    $P(Z  <  -0.71) =  0.2388$

So

     $P( \^ p  <  0.35)=    0.2388$