Question

The time it takes a randomly selected rat to complete a maze is normally distributed with mean 1.5 minutes and standard deviation 0.35 minutes.

(a) Find the probability that the completion time from a randomly selected rat is shorter than 1.4 minutes. (show work)

(b) Find the probability that the average time from 4 randomly selected rats is shorter than 1.4 minutes. (show work)

(c) Find the probability that the average time from 100 randomly selected rats is shorter than 1.4 minutes (show work)

Answer 1

Answer:

a)

Given,

Mean = 1.5

Standard deviation = 0.35

P(X < 1.4) = P((x-u)/s < (1.4 - 1.5)/.35)

= P(z < -0.29)

= 0.3859081 [since from z table]

= 0.3859

b)

P(X < 1.4) = P((x-u)/(s/sqrt(n)) < (1.4 - 1.5)/(0.35/sqrt(4)))

= P(z < -0.57)

= 0.2843388 [since from z table]

= 0.2843

c)

P(X < 1.4) = P((x-u)/(s/sqrt(n)) < (1.4 - 1.5)/(0.35/sqrt(100)))

= P(z < -2.86)

= 0.0021182 [since from z table]

= 0.0021