Solve the problem.
-Human body temperatures are normally distributed with a mean of 98.20°F and a standard deviation of 0.62°F. If 19 people are randomly selected, find the probability that their mean body temperature will be less than 98.50°F.
A)0.3343
B)0.4826
C)0.9826
D)0.0833

Question

Quizplus · Accepted Answer

Using the formula for the standard error of the mean, which is the standard deviation divided by the square root of the sample size (n), we get 0.62 / sqrt(19). To find the Z-score for 98.50°F, we use the formula Z = (X - μ) / (σ/√n), where X is 98.50, μ is 98.20, and σ/√n is the standard error calculated. Plugging in the values, we find the Z-score for 98.50°F. Using the Z-score, we look up the corresponding probability in the standard normal distribution table, which gives us the probability of the mean body temperature being less than 98.50°F for a sample of 19 people. This calculation results in a probability of 0.9826, indicating a high likelihood that the mean temperature of a randomly selected group of 19 people will be less than 98.50°F.

Solve the Problem. -Human Body Temperatures Are Normally Distributed with a Mean of a Mean