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 65 healthy adults, find the probability that the average body temperature is between 98.43 and 98.73. (Correct to 4 decimal places)

Answer 1

solurion=

Given that ,

mean =   = 98.6

standard deviation = = 0.6  

n = 65

= 98.6

=  / n= 0.6 / 65=0.0744

P(98.43<     <98.73 ) = P[(98.43-98.6) /0.0744  < ( - ) /   < (98.73-98.6) /0.0744 )]

= P( -2.28< Z <1.75 )

= P(Z <1.75 ) - P(Z < -2.28)

Using z table

=0.9599-0.0113

=0.9486

probability= 0.9486