In: Statistics and Probability
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,
A) find the proportion of mice that live more than 55 months
B) find the proportion of mice that live between 40 and 52 months
C) Find the 99th percentile of the lifetimes of such mice
Answer:
Given,
Mean = 40
Standard deviation = 6.3
a)
To determine the proportion of mice that live more than 55 months
P(X > 55) = 1 - P(X < 55)
= 1 - P((x-mu)/s < (55 - 40)/6.3)
= 1 - P(z < 2.38)
= 1 - 0.9913436 [since from standard normal distribution table]
= 0.0087
Hence the proportion of mice that live more than 55 months is 0.0087
b)
To give the proportion of mice that live between 40 and 52 months
P(40 < X < 52) = P((40 - 40)/6.3) < (x - mu)/s < (52 - 40)/6.3)
= P(0 < z < 1.90)
= P(z < 1.90) - P(z < 0)
= 0.9712834 - 0.5 [since from standard normal distribution table]
= 0.4713
Hence the proportion of mice that live between 40 and 52 months is 0.4713
c)
To give the 99th percentile of the lifetimes of such mice
P(X < x) = 0.99
P(z < (x - 40)/6.3) = 0.99
(x - 40)/6.3 = 2.33 [since from standard normal distribution table]
x - 40 = 2.33*6.3
x = 40 + 14.676
x = 54.679