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 standard deviation of 6.3 month, 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

Answer:

answer is there

ok see there

Answer 2

Answer:

Probability that a mice will live more than 32 months = 0.8980

Probability that a mice will live less than 28 months = 0.0287

Probability that a mice will live between 37 and 49 months = 0.608

Step-by-step explanation:

Given:

Mean, μ = 40

Standard deviation , = 6.3

a) Probability that mice will live more than 32 months

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

Considering the z table, the probability that the mice will live more than 32 months is :

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

b)Probability that mice will live less than 28 months

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

Considering the z table, the probability that the mice will live less than 28 months is :

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

               =1-0.9713

                =0.0287

c)Probability that mice will live between 37 months and 49 months

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

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

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

                               =0.6844+0.9236-1

                                =0.608

Hence,

Probability that the mice will live more than 32 months = 0.897  Probability that the mice will live less than 28 months = 0.0287 Probability that the mice will live between 37 and 49 months =   0.608