Using the Normal Distribution Find the temperature of a person in percentile-71, given that we expect the average temperature to be 98.6 with a standard deviation of 0.7. Round your answer to one decimal place. percentile 71 + 96.5 97.2 97.9 98.6 99.3 100 100.7 101. temp Q temperature = Hint: Try using StatKey _ or a spreadsheet, =NORMINV(p, mean, std)