Question

The time it takes a randomly selected rat of a certain subspecies to find its way through a maze is normally distributed with µ = 1.5 minutes and σ = 0.35 minutes. Suppose 30 rats are selected. Let X1,X2,...,X30 denote their times in the maze.

(a) What are the mean and standard deviation for total time for all 30 rats in the maze?

(b) What is the probability that the total time for the 30 rats is between 45 and 47 minutes?

(c) What is the probability that the sample mean maze time for the rats is between 1.1 and 1.6 minutes?

(d) Find the 95th percentile for the sample mean time. (NOTE: This is equivalent to asking to find the cutoff for the slowest 5% of the rats in the sample.

We usually say the 95% percentile is the upper 5%. Here, we are looking at the upper 5% of maze times. Longer time = slower rat)

Answer 1

Given : µ = 1.5 minutes and σ = 0.35 minutes

a) According to sampling distribution of sample mean. The sample mean approximately follows normal distribution with mean = and standard deviation =

Mean for total time for all 30 rats in the maze () = 1.5

Standard deviation for total time for all 30 rats in the maze () = 0.35/√ 30 = 0.0639

b) What is the probability that the total time for the 30 rats is between 45 and 47 minutes?

P( 45/30 ≤ ≤ 47/30 ) = P( 1.5 ≤ ≤ 1.567 )

=P( ≤ 1.567 ) - P( ​​​​​​​ ≤ 1.5 )

=

= P( z ≤ 1.04 ) - P( z ≤ 0 )

= 0.8508 - 0.5000 ---- ( from z score table )

=0.3508

Probability that the total time for the 30 rats is between 45 and 47 minutes is 0.3508

c) What is the probability that the sample mean maze time for the rats is between 1.1 and 1.6 minutes?

P( 1.1 ≤ ≤ 1.6 )

= P( ​​​​​​​ ≤ 1.6 ) - P( ​​​​​​​ ≤ 1.1 )

=

= P( z ≤ 1.56 ) - P( z ≤ -6.25 )

= 0.9406 - 0 ---- ( from z score table )

=0.9406

The probability that the sample mean maze time for the rats is between 1.1 and 1.6 minutes is 0.9406

(d) Find the 95th percentile for the sample mean time

P(   ≤ ) = 0.95

We have to find z score corresponding to area 0.9500 on the z score table

So z = 1.645 ---- ( from z score table )

= ( Z* ) + = ( 1.645*0.0639) + 1.5

= 1.61

The 95th percentile for the sample mean time is 1.61 minutes.