Question

A Snapshot in USA Today indicates that 51% of Americans say the average person is not very considerate of others when talking on a cellphone. Suppose that 100 Americans are randomly selected. Find the approximate probability that 52 or more Americans would indicate that the average person is not very considerate of others when talking on a cellphone. (Use the normal approximation. Round your answer to four decimal places.)

Answer 1

Answer:

The approximate probability is  $P(X > 0.52) = P(Z > 0.2 ) = 0.42074$

Step-by-step explanation:

From the question we are told that

  The population proportion of Americans  that say the average person is not considerate of others when talking on a cellphone  is  p =  0.51

   The sample size is  n  =  100

Generally because the sample size is sufficiently  large the mean of this sampling distribution is

       $\mu_{x} = p = 0.51$

Generally the standard deviation is mathematically represented as

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

=> $\sigma = \sqrt{ \frac{ 0.51 (1 - 0.51)}{100} }$

=> $\sigma = 0.04999$

Generally  the sample proportion is mathematically represented as

        $\^ p = \frac{52}{100}$

=>     $\^ p = 0.52$

Generally the probability that  52 or more Americans would indicate that the average person is not very considerate of others when talking on a cellphone is mathematically represented as

     $P(X > 0.52) = P( \frac{X - \mu_{x}}{\sigma} > \frac{ 0.52 - 0.51}{0.04999} )$

$\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )$

=>   $P(X > 0.52) = P(Z > 0.2 )$

From the z table  the area under the normal curve corresponding to  0.2 to the right is  

      $P(X > 0.52) = P(Z > 0.2 ) = 0.42074$