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Question
Statistics
Posted 8 months ago

The amount of cleaning solution a company fills its bottles with has a mean of 33floz33 \mathrm{fl} \mathrm{oz} and a standard deviation of 1.5floz1.5 \mathrm{fl} \mathrm{oz}. The company advertises that these bottles have 32floz32 \mathrm{fl} \mathrm{oz} of cleaning solution.

What will be the mean and standard deviation of the distribution of excess cleaning solution, in milliliters?
(1floz(1 \mathrm{fl} \mathrm{oz} is approximately 30 mL30 \mathrm{~mL}.
Choose 1 answer:
(A) Mean: 30 mL30 \mathrm{~mL}
Standard deviation: 13 mL13 \mathrm{~mL}
(B) Mean: 30 mL30 \mathrm{~mL}
Standard deviation: 45 mL45 \mathrm{~mL}
(C) Mean: 958 mL958 \mathrm{~mL}
Standard deviation: 13 mL13 \mathrm{~mL}
(D) Mean: 958 mL958 \mathrm{~mL}
Standard deviation: 45 mL45 \mathrm{~mL}
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Answer from Sia
Posted 8 months ago
Solution by Steps
step 1
Convert the mean and standard deviation from fluid ounces to milliliters using the conversion factor 1 fl oz=30 mL1 \text{ fl oz} = 30 \text{ mL}
step 2
Calculate the mean of the excess cleaning solution in milliliters: Mean in mL=(33 fl oz32 fl oz)×30 mL/fl oz=1 fl oz×30 mL/fl oz=30 mL \text{Mean in mL} = (33 \text{ fl oz} - 32 \text{ fl oz}) \times 30 \text{ mL/fl oz} = 1 \text{ fl oz} \times 30 \text{ mL/fl oz} = 30 \text{ mL}
step 3
Calculate the standard deviation of the excess cleaning solution in milliliters: Standard deviation in mL=1.5 fl oz×30 mL/fl oz=45 mL \text{Standard deviation in mL} = 1.5 \text{ fl oz} \times 30 \text{ mL/fl oz} = 45 \text{ mL}
Answer
D
Key Concept
Conversion of units and calculation of mean and standard deviation
Explanation
To find the mean and standard deviation of the excess cleaning solution in milliliters, we first convert the given values from fluid ounces to milliliters using the provided conversion factor.

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