Answered

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A large random sample was taken of body temperatures of women at a university. The data from the sample were normally distributed with a mean body temperature of
98.52°F and a standard deviation of 0.727°F. Based on this sample, which percentage is the best estimate of the proportion of all women at this university who have a body
temperature more than 2 standard deviations above the mean?
O 0.30%
O 2.28%
72.70%
O 97.72%

Sagot :

oladayoademosu

The best estimate of the proportion of all women at this university who have a body temperature more than 2 standard deviations above the mean is 2.28%.

Since the sample is normally distributed and has a mean μ = 98.52 F and a standard deviation, σ = 0.727 F and we need to find the percentage of all womenat this university who have a body temperature more than 2 standards deviations above the mean.

The normal distribution

Since the sample is normally distributed, 50% of the sample is below the mean. We have 34% of the sample at 1 standard deviation away from the mean and 47¹/₂% at 2 standard deviations above the mean.

Percentage below 2 standard deviations away from mean

So, the percentage below 2 standard deviations from the mean is 50% + 47¹/₂% = 97¹/₂%.

Percentage above 2 standard deviations away from mean

So, the percentage above 2 standard deviations from the mean is 100% - 97¹/₂% = 2¹/₂% = 2.5 %

Since 2.28 % is the closest to 2.5 % from the options, the best estimate is 2.28 %.

So, the best estimate of the proportion of all women at this university who have a body temperature more than 2 standard deviations above the mean is 2.28%.

Learn more about normal distribution here:

https://brainly.com/question/25800303

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