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 98.242 and a standard deviation of 0.713.

1) What is the probability that a randomly selected person's body temperature is between 97.9 and 98.51? Round your answer to 4 decimal places.
2) What is the probability that the average body temperature of 4 randomly selected people is between 97.9 and 98.51? Round your answer to 4 decimal places.

3) Why did the probability increase?

• The probability increased since the sample size decreased resulting in a wider distribution which increased the chances of being in an interval containing the mean.
• The probability increased since the sample size increased resulting in a narrower distribution which increased the chances of being in an interval containing the mean.

Answer 1

Solution :

Given that ,

mean = = 98.242

standard deviation = = 0.713

1)P( 97.9< x < 98.51 ) = P[(97.9 -98.242)/0.713 ) < (x - ) /  < (98.51 - 98.242) /0.713 ) ]

= P( -0.48< z < 0.38)

= P(z < 0.38) - P(z < -0.48 )

Using standard normal table

= 0.648 - 0.3156= 0.3324

Probability = 0.3324

2)

n = 4

= = 98.242

= / n = 0.713 / 4 = 0.3565

P(97.9< < 98.51 )  

= P[(97.9 - 98.242) /0.3565 < ( - ) / < (98.51 - 98.242) /0.3565 )]

= P( -0.96< Z < 0.75 )

= P(Z < 0.75 ) - P(Z <-0.96 )

= 0.7734 - 0.1685 = 0.6049

probability = 0.6049

c) The probability increased since the sample size increased resulting in a narrower distribution which increased the chances of being in an interval containing the mean.