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The amount of cleaning solution a company fills its bottles with has a mean of 33 fl oz and a standard deviation of 1.5 fl oz. The company advertises that these bottles have 32 fl oz of cleaning solution. What will be the mean and standard deviation of the distribution of excess cleaning solution, in milliliters? (1 fl oz is approximately 30 mL.)

Choose 1 answer:

A. Mean: 30 mL, Standard deviation: 13 mL
B. Mean: 30 mL, Standard deviation: 45 mL
C. Mean: 958 mL, Standard deviation: 13 mL

Answer :

Final answer:

The company's excess cleaning solution has a mean of 30 mL and a standard deviation of 45 mL.

Explanation:

The student's question involves translating the statistics of a distribution from fluid ounces (fl oz) to milliliters (mL), keeping in mind that 1 fl oz is approximately equal to 30 mL. The company's cleaning solution bottles have a mean of 33 fl oz and a standard deviation of 1.5 fl oz. But they advertise as holding only 32 fl oz. So, the excess amount that the company fills on average is 33 fl oz - 32 fl oz = 1 fl oz.

After converting this to milliliters, the average excess cleaning solution would be 1 fl oz * 30 mL = 30 mL.

In terms of standard deviation, we are just converting units from one to another. So, the original standard deviation (1.5 fl oz) would convert to 1.5 fl oz * 30 mL = 45 mL in mL. Therefore, the mean and standard deviation of the distribution of excess cleaning solution is 30 mL and 45 mL respectively.

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