Question

what is the approximate probability that in a random sample of 1000 individuals who have purchased fries at McDonald’s, at least 40 can taste the difference between the two oils? (2.5 pts.)

Answer 1

Complete Question

In response to concerns about nutritional contents of fast foods, McDonald's has announced that it will use a new cooking oil for its French fries that will decrease substantially trans fatty acid levels and increase the amount of more beneficial polyunsaturated fat. The company claims that only 3 out of 100 people can detect a difference in taste between the new and old oils. Assuming that this figure is correct (as a long-run proportion), what is the approximate probability that in a random sample of 1000 individuals who have purchased fries at McDonald's, at least 40 can taste the difference between the two oils? (2.5 pts.)

Answer:

The probability is  $P(\^ p \ge 0.04 ) = 0.03216$

Step-by-step explanation:

From the question we are told that

   The population proportion is  $p = \frac{3}{100} = 0.03$

    The sample size is  n  = 1000

 Generally the mean of this sampling distribution is  $\mu_{x} = 0.03$

Generally the standard deviation of this sapling distribution is  mathematically evaluated as  

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

=>      $\sigma = \sqrt{\frac{0.03 (1 - 0.03)}{1000} }$

=>      $\sigma = 0.0054$

Generally the sample proportion when the number of those that can taste the difference is 40  is mathematically represented as

      $\^ p = \frac{40}{1000} = 0.04$

Generally the approximate probability that in a random sample of 1000 individuals who have purchased fries at McDonald's, at least 40( $\^ p = 0.04$) can taste the difference between the two oils is mathematically represented  as

     $P(\^ p \ge 0.04 ) = 1 - P(\^ p < 0.04 )$

Here

      $P(\^ p < 0.04 ) = P(\frac{ \^ p - \mu_{x}}{\sigma } < \frac{0.04 - 0.03}{0.0054} )$

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

=>   $P(\^ p < 0.04 ) = P(Z < 1.85 )$

From the z table the probability of  (Z < 1.85  ) is

      $P(Z < 1.85 ) = 0.96784$

So

    $P(\^ p < 0.04 ) = 0.96784$

So

   $P(\^ p \ge 0.04 ) = 1 - 0.96784$

=> $P(\^ p \ge 0.04 ) = 0.03216$