Question

Fast Food and Gas Stations Forty percent of all Americans who travel by car look for gas stations and food outlets that are close to or visible from the highway. Suppose a random sample of n = 25 Americans who travel by car are asked how they determine where to stop for food and gas. Let x be the number in the sample who respond that they look for gas stations and food outlets that are close to or visible from the highway.
a. What are the mean and variance of x?
b. Calculate the interval p-±2a. What values of the binomial random variable x fall into this interval?
c. Find P(6 ≤ x ≤ 14). How does this compare with the fraction in the interval WO for any distribution? For mound-shaped distributions?

Answer 1

Answer:

a

mean $\mu = 10$ variance $\sigma^2 = 6$

b

The binomial random variable x fall into this interval ranges from

     - 5  to  5

c

$P(6 \le x \le 14) = 0.8969$  

Step-by-step explanation:

From the question we are told that

  The sample size is  $n = 25$

   The  percentage that look for gas stations and food outlets that are close to or visible from the highway is  $p = 0.40$

Generally the mean is mathematically represented as

       $\mu = n * p$

=>     $\mu = 0.40 * 25$

=>    $\mu = 10$

The variance is mathematically represented as

      $\sigma^2 = np(1- p )$

=>   $\sigma^2 = 25 * 0.40(1- 0.40 )$

=>    $\sigma^2 = 6$

The  standard deviation is mathematically evaluated as

     $\sigma = \sqrt{\sigma^2}$

     $\sigma = \sqrt{6}$

    $\sigma = 2.45$

The  interval is evaluated as

       $p\pm 2 \sigma$

=>   $p - 2 \sigma\ \ \ , \ \ \ p + 2\sigma$

=>   $0.40 - 2 *2.45\ \ \ , \ \ \ 0.40 + 2* 2.45$

=>   $-4.5\ \ \ , \ \ \ 5.3$

The binomial random variable x fall into this interval ranges from

     - 5  to  5

Generally

    $P(6 \le x \le 14) = P(\frac{ x - \mu }{\sigma } \le \frac{14 - 10}{{2.45}} ]-P[ \frac{ x - \mu }{\sigma } \le \frac{6 - 10}{2.45 } ]$

     $P(6 \le x \le 14) = P(Z \le 1.63 ]-P[ Z \le -1.63 ]$

     $P(6 \le x \le 14) = [1- P(Z > 1.63 ]] -[1- P[ Z > -1.63 ]]$      

From the z-table  

                 $P(Z > 1.63 ) = 0.051551$

And

                $P(Z >- 1.63 ) =0.94845$

=>        $P(6 \le x \le 14) = [1-0.051551] -[1-0.94845]$      

=>        $P(6 \le x \le 14) = 0.8969$