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The body temperature of adults are normally distributed with a mean of 98.6 f and a standard deviation of 0.71f what temperature represents the 95th percentile ?

User PeterSW
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2 Answers

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Final answer:

The temperature that represents the 95th percentile for an adult's body temperature, normally distributed with a mean of 98.6°F and a standard deviation of 0.71°F, is approximately 99.77°F.

Step-by-step explanation:

To find the temperature that represents the 95th percentile for a normally distributed variable (like body temperature), you can use the Z-score table or a statistical software that can provide the percentile based on the mean and standard deviation. Since the mean body temperature is given as 98.6°F with a standard deviation of 0.71°F for adults, we will use the Z-score formula to find the 95th percentile.

Firstly, look up the Z-score that corresponds to the 95th percentile in the Z-table, which is approximately 1.645. Then, use the Z-score formula:

Z = (X - mean) / standard deviation

Rearrange the formula to solve for X, which represents the temperature at the 95th percentile:

X = Z * standard deviation + mean

Substitute the values into the formula:

X = 1.645 * 0.71 + 98.6

X = 1.16815 + 98.6

X = 99.76815°F

Therefore, the temperature that represents the 95th percentile is approximately 99.77°F.

User Siddiqui Noor
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User MarkDBlackwell
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