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.
<h3>
The normal distribution</h3>
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.
<h3>Percentage below 2 standard deviations away from mean</h3>
So, the percentage below 2 standard deviations from the mean is 50% + 47¹/₂% = 97¹/₂%.
<h3>Percentage above 2 standard deviations away from mean</h3>
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:
brainly.com/question/25800303