Question

A company is planning to survey the passengers on a particular bus service. Over the years, the average number of passengers per trip on the bus has been 60 and the standard deviation of the number of passengers per trip has been 18.
q13. If 36 bus trips were randomly selected, what is the probability that the average number of passengers (for the sample of 36 trips) would be more than 66 ie. P X( 66) > ?
(a) 0.0228
(b) 0.4772
(c) 0.3694
(d) 0.1306
(e) cannot be determined unless the number of passengers per trip is normally distributed

Q14. If 9 Bus trips were randomly selected, what is the probability that the average number of passengers (for the sample of 9 trips) would be between 54 and 66 ie. P X (54 66) < < ?
(a) 0.6826
(b) 0.2611
(c) 0.3413
(d) 0.9545
(e) cannot be determined unless the number of passengers per trip is normally distributed

Q15. If 9 bus trips were randomly selected, then the standard error of the mean ( X σ ) for the variable X, the number of passengers per trip, is (a) 0.1111
(b) 18
(c) 2
(d) 6
(e) Cannot be determined as the sample size is too small (n < 30)

Answer 1

It is given that : The average number of passengers per trip on the bus has been 60 and the standard deviation of the number of passengers per trip has been 18.

Let X = The number of passengers per trip on the bus.

So   

and

q13. If 36 bus trips were randomly selected, what is the probability that the average number of passengers (for the sample of 36 trips) would be more than 66 ie. P X( 66) > ?

Here n = sample size is greater than 30 so we can used central limit theorem ( CLT )

The sampling distribution of the sample mean is approximately normally distributed with mean and standard deviation as follows:

Here we want to find P( > 66) = 1 - P( < 66) .....( 1 )

Using excel:

P( < 66) = "=NORMDIST(66,60,3,1)" = 0.9772

Plug this value in equation ( 1 ), we get:

P( < 66) = 1 - 0.9772 = 0.0228

So correct choice is a)

Q14. If 9 Bus trips were randomly selected, what is the probability that the average number of passengers (for the sample of 9 trips) would be between 54 and 66 ie. P X (54 66) < < ?

Here n = 9

Therefore

We want to find

P( < 66) = "=NORMDIST(66,60,6,1)" = 0.8413

And P( < 54) = "=NORMDIST(54,60,6,1)" = 0.1587

Plug these values in equation 1) we, get

P( 54 < < 66) = 0.8413 - 0.1587 = 0.6826

So correct choice is ( a )

Q15. If 9 bus trips were randomly selected, then the standard error of the mean ( X σ ) for the variable X, the number of passengers per trip, is

From the above parts the standard deviation of sample mean is nothing but the standard error of the mean .

Therefore the standard error =

So correct choice is ( d ) 6.