Question

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Suppose Company A produces packages of throat lozenges that are normally distributed with a mean of 38.2 individual lozenges and a standard deviation of 1.7 lozenges.

Company B produces packages of throat lozenges that are normally distributed with a mean of 36.9 individual lozenges and a standard deviation of 2.4 lozenges.

Select from the drop-down menus to correctly complete the statement.

CHOICES ARE COMPANY A AND COMPANY B

Answer 1

Answer:  B, A, B

Step-by-step explanation:

Let's find the z-score for both companies using the formula: $z=\dfrac{x-\mu}{\sigma}$

Company A: $z=\dfrac{43-38.2}{1.7}=\large\boxed{2.8}$

Company B: $z=\dfrac{43-36.9}{2.4}=\large\boxed{2.5}$

The z-score from Company B is closer to 0 than the z-score of Company A so it it more likely that a package of 43 lozenges is produced by Company B

Answer 2

Answer: B, A and B.

Step-by-step explanation: I took the quiz.

Screenshot provided

Since the z-score from Company B is closer to zero than the z-score of Company A, the company that produced a package with 43 lozenges in it is most likely Company B.