Answer to Question #106986 in Statistics and Probability for Putra

Question

A study states that mice will have an average age of 40 months when given a certain diet and enriched with additional vitamins and protein. Assume that the life span of the mice is normally distributed with a standard deviation of 6.3 months. Determine the chances that the mouse will have a lifetime: a. More than 32 months b. Less than 28 months c. Between 37 and 49 months

Answer 1

Mean =30 months

Sd=6.3 months

Let x be the random variable for number of months

a) P(X>32)

Z=$\frac {X-mean} {sd}$

=$\frac {32-40} {6.3}$

=-1.27

P(z>-1.27)=P(z<1.27) from the z tables is

0.89796

b) p(X <28)

Z=$\frac {28-40} {6.3}$

=-1.90

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

P(z<1.90) from the z-tables is 0.97128

P(z<-1.90)=1-0.97128

=0.02872

c) P(37<X<49)

Z1=$\frac {37-40} {6.3}$

=-0.48

Z2=$\frac {49-40} {6.3}$

=1.43

P(-0.48<z<1.43)=P(Z1<0.48)+

P(Z2<1.43)-1

The values are obtained from the z tables

=0.68439+0.92364-1

=0.60803