Answer to Question #292027 in Statistics and Probability for loop

Question

According to the article "A Critical Appraisal of 98.6 Degrees F, the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August Wunderlich" published in the Journal of the American Medical Association, the body temperatures of adults are normally distributed with a mean of 98.261 and a standard deviation of 0.733.

1. What is the probability that a randomly selected person's body temperature is between 98.02 and 98.53? Round your answer to 4 decimal places.
2. What is the probability that the average body temperature of 7 randomly selected people is between 98.02 and 98.53? Round your answer to 4 decimal places.

Answer 1

1) $P(98.02<X<98.53)=P(\frac{98.02-98.261}{0.733}<Z<\frac{98.53-98.261}{0.733})=$

$=P(-0.33<Z<0.37)=P(Z<0.37)-P(Z<-0.33)=0.2736.$

2) $P(98.02<\bar X<98.53)=P(\frac{98.02-98.261}{\frac{0.733}{\sqrt{7}}}<Z<\frac{98.53-98.261}{\frac{0.733}{\sqrt{7}}})=$

$=P(-0.87<Z<0.97)=P(Z<0.97)-P(Z<-0.87)=0.6418.$