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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 2.2 lozenges.

Which company would be more likely to produce a package of 43 throat lozenges?
Company A is more likely; its z-score of -2.82 is closer to the mean than Company B's z-score of -2.77.

Company B is more likely; its z-score of -2.77 is closer to the mean than Company A's z-score of -2.82.

Company A is more likely; its z-score of 2.82 is closer to the mean than Company B's z-score of 2.77.

Company B is more likely; its z-score of 2.77 is closer to the mean than Company A's z-score of 2.82.

Respuesta :

Answer:

The correct option is:

Company B is more likely; its z-score of 2.77 is closer to the mean than Company A's z-score of 2.82.

Explanation:

The company's whose z score would be closer to 0 would be more likely to produce the package of 43 throat lozenges

The z score for the company A is:

[tex]z=\frac{43-38.2}{1.7} =2.82[/tex]

While the z score for the company B is:

[tex]z=\frac{43-36.9}{2.2}=2.77[/tex]

Since the z score of company B is closer to 0 than the company A, therefore the company is more likely

xxkidtyrantxx

Answer:

B,A,B

Step-by-step explanation:

>:) Did The Test

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