The time spent in days waiting for a kidney transplant for p

The time spent (in days) waiting for a kidney transplant for people ages 35-49 can be approximiated by the normal distribution, as shown in the figure to the right. (a) What waiting time represents the 98th percentile? (b) What waiting time represents the first quartile? mean of1676 and standard deviation of 211.6

Solution

Given mean =1676; standard deviation = 211.6

98th percentile is P(z<z*) =0.98

=P(z<2.054)=0.98

98the percentile is = 2.054*211.6 + 1676 ( since z=( x- mean)/sd)

=2110.573

first quarile is P(z<z*) =0.25

from standard normal tables P(Z<-0.67499) =0.25

first quartile value is = -0.67499*211.6 + 1676 = 1533.278