Question: Body temperatures of adults are normally distributed with a mean of 98.6°F and a standard deviation…

Body temperatures of adults are normally distributed with a mean of 98.6°F and a standard deviation of 0.6°F.

39. What percentage of adults have body temperatures less than 98.6°F?

40. What percentage of adults have body temperatures less than 97°F?

41. What percentage of adults have body temperatures greater than 97°F?

42. What percentage of adults have body temperatures greater than 100°F?

43. What percentage of adults have body temperatures within one standard deviation of the mean?

44. What percentage of adults have body temperatures between 98°F and 99°F?

45. What is your body temperature if it is more than that of 80% of the population?

46. What is your body temperature if it is less than that of 80% of the population?

47. What is your body temperature if it is more than that of 99% of the population?

Solution

The question involves a normally distributed data set of body temperatures with a mean of 98.6°F and a standard deviation of 0.6°F. Let’s solve the problems step by step using the properties of the normal distribution. #### 39) What percentage of adults have body temperatures less than 98.6°F? Since 98.6°F is the mean of the distribution, 50% of adults will have body temperatures less than 98.6°F. #### 40) What percentage of adults have body temperatures less than 97°F? First, calculate the z-score: \[ z = \frac{97 - 98.6}{0.6} = \frac{-1.6}{0.6} = -2.67 \] Using a standard normal distribution table, find the percentage corresponding to a z-score of -2.67. This is approximately 0.38%. #### 41) What percentage of adults have body temperatures greater than 97°F? Find the percentage greater than 97°F: Since 0.38% have temperatures less than 97°F, \[ 100\% - 0.38\% = 99.62\% \] #### 42) What percentage of adults have body temperatures greater than 100°F? Calculate the z-score: \[ z = \frac{100 - 98.6}{0.6} = \frac{1.4}{0.6} = 2.33 \] Find the percentage: A z-score of 2.33 corresponds to 99.01% below it, so \[ 100\% - 99.01\% = 0.99\% \] #### 43) What percentage of adults have body temperatures within one standard deviation of the mean? This corresponds to the range 98.0°F to 99.2°F. Calculate z-scores for 98.0 and 99.2: Lower bound: \[ z = \frac{98.0 - 98.6}{0.6} = -1 \] Upper bound: \[ z = \frac{99.2 - 98.6}{0.6} = 1 \] The percentage of data within one standard deviation in a normal distribution is approximately 68%.