In: Statistics and Probability
The body temperatures of adults are normally distributed with a mean of 98.6° F and a standard deviation of 0.50° F. If 25 adults are randomly selected, find the probability that their mean body temperature is greater than 98.4° F.
Solution :
Given that,
mean = = 98.60 F
standard deviation =
= 0.500 F
n = 25
=
= 98.60 F
=
/
n = 0.50 /
25 = 0.10
P(
> 98.40 F) = 1 - P(
< 98.40 F)
= 1 - P[(
-
) /
< (98.4 - 98.6) / 0.10]
= 1 - P(z < -2.00)
Using z table,
= 1 - 0.0228
= 0.9772