Answer to Question #256806 in Statistics and Probability for Chiran

Question

A research scientist reports that mice will live an average of 40 months when their diets are sharply restricted and then enriched with vitamins and proteins. Assuming that the lifetimes of such mice are normally distributed with a standard deviation of 6.3 months, find the probability that a given mouse will live (a) more than 32 months; (b) less than 28 months; (c) between 37 and 49 months.

Answer 1

Mean = 40

Standard deviation = 6.3

(a) more than 32 months:

$P(X>32)=\frac{X-mean}{\sigma}=\frac{32-40}{6.3}=-1.27$

From z table probability of more than 32 months is:

$P(z>-1.27)=1-0.1021=0.8979$

(b) less than 28 months:

$P(X<28)=\frac{28-40}{6.3}=-1.90$

From z table probability of less than 28 months is:

$P(z<-1.90)=1-P(z<1.90)$

$P(z<-1.90)=1-0.9713=0.0287$

(c) between 37 and 49 months:

$P(37<X<49)=(\frac{37-40}{6.3}<z<\frac{49-40}{6.3})$

$P(37<X<49)=(-0.48<z<1.43)$

$P(-0.48<z<1.43)=P(z<0.48)+P(z<1.43)-1$

$P(-0.48<z<1.43)=0.6844+0.9236-1=0.608$