Question

An 1868 paper by German physician Carl Wunderlich reported, based on over a million body temperature readings, that healthy adult body temperatures are approximately normally distributed with mean 98.6 degrees Farenheit and standard deviation 0.6. In a random sample of 40 healthy adults, find the probability that the average body temperature is between 98.43 and 98.74. (4 decimal places)

Answer 1

Solution :

Given that ,

mean =   = 98.6

standard deviation = = 0.6

n = 40

= 98.6

=  / n= 0.6 / 40=0.09486

P(98.43<     < 98.74) = P[(98.43-98.6) / 0.09486 < ( - ) /   < (98.74-98.6) /0.09486 )]

= P(-1.79 < Z <1.48 )

= P(Z <1.48 ) - P(Z < -1.79)

Using z table

=0.9306-0.0367

=0.8939

probability=0.8939  

Solution