Question

Suppose the time between buses at a particular stop is a positively skewed random variable with an average of 60 minutes and standard deviation of 6 minutes. Suppose the time between buses at this stop is measured for a randomly selected week, resulting in a random sample of n = 36 times. The average of this sample,

X, is a random variable that comes from a specific probability distribution.

(a)Which of the following is true about the distribution of mean times for n = 36?

The distribution will be normally distributed with a mean of 60 minutes and a standard deviation of 6 minutes.

The distribution will be positively skewed with a mean of 60 minutes and a standard deviation of 6 minutes.   

The distribution will be normally distributed with a mean of 60 minutes and a standard deviation of 1 minutes.

The distribution will be positively skewed with a mean of 60 minutes and a standard deviation of 1 minutes.

(b) Calculate the probability the mean time for the sample of 36 buses will be between 59 minutes and 61 minutes.

P(59 ≤ X ≤ 61)

=  

(c) How likely is it the average time will exceeds 61 minutes?

P(X ≥ 61)

=

You may need to use the z table to complete this problem.

Answer 1

(a)Which of the following is true about the distribution of mean times for n = 36?

The distribution will be normally distributed with a mean of 60 minutes and a standard deviation of 1 minutes.

For the mean times the distribution follows Normal distribution with mean = 60,

(b) Calculate the probability the mean time for the sample of 36 buses will be between 59 minutes and 61 minutes.

                               
P ( 59 < X < 61 )                                   
Standardizing the value                                   
                           
                       
Z = -1                                  
                               
Z = 1                                  
P ( -1 < Z < 1 )                                   
P ( 59 < X < 61 ) = P ( Z < 1 ) - P ( Z < -1 )                                   
P ( 59 < X < 61 ) = 0.8413 - 0.1587                                  
P ( 59 < X < 61 ) = 0.6827  

(c) How likely is it the average time will exceeds 61 minutes?

                           
P ( X > 61 ) = 1 - P ( X < 61 )                                   
Standardizing the value                                   
                                  
                              
Z = 1                                  
                       
P ( Z > 1 )                                   
P ( X > 61 ) = 1 - P ( Z < 1 )                                   
P ( X > 61 ) = 1 - 0.8413                                  
P ( X > 61 ) = 0.1587