A study of human body temperatures using healthy women showe

A study of human body temperatures using healthy women showed a mean of 98.4*f and a standard deviation of about .70*f, assume the temperatures are approximately normally distributed

A) Find the percentage of healthy women with temperatures below 98.6*f

B) what temperature does healthy woman have if her temperature is at the 76th percentile?

Solution

Normal Distribution
Mean ( u ) =98.4
Standard Deviation ( sd )=0.7
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X < 98.6) = (98.6-98.4)/0.7
= 0.2/0.7= 0.2857
= P ( Z <0.2857) From Standard Normal Table
= 0.6125                  
b)
P ( Z < x ) = 0.76
Value of z to the cumulative probability of 0.76 from normal table is 0.706
P( x-u/s.d < x - 98.4/0.7 ) = 0.76
That is, ( x - 98.4/0.7 ) = 0.71
--> x = 0.71 * 0.7 + 98.4 = 98.8942