Question

The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.2 minutes and standard deviation 1.5 minutes. Suppose that a random sample of n = 49 customers is observed. Find the probability that the average waiting time in line for these customers is

Answer 1

Part a

`P(\bar X <10)=P(Z<(10-8.2)/((1.5)/(√(49)))=8.4)`

And using a calculator, excel or the normal standard table we have that:

`P(Z<8.4)=1`

Part b

`P(5<\bar X <10)=P((5-8.2)/((1.5)/(√(49)))<Z<(10-8.2)/((1.5)/(√(49)))) =P(-14.933< Z<8.4)`

And using a calculator, excel or the normal standard table we have that:

`P(-14.933<Z<8.4)=1-0 = 1`

Part c

`P(\bar X <6)=P(Z<(6-8.2)/((1.5)/(√(49)))=-10.267)`

And using a calculator, excel or the normal standard table we have that:

`P(Z<-10.267)=0`

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the time waiting at an airport check in of a population, and for this case we know the distribution for X is given by:

`X \sim N(8.2,1.5)`

Where
`\mu=8.2` and
`\sigma=1.5`

We select a random sample of n = 49. Since the distribution of X is normal then the distribution for the sample mean
`\bar X` is also normal and given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

We can find the individual probabilities like this:

Part a

`P(\bar X <10)=P(Z<(10-8.2)/((1.5)/(√(49)))=8.4)`

And using a calculator, excel or the normal standard table we have that:

`P(Z<8.4)=1`

Part b

`P(5<\bar X <10)=P((5-8.2)/((1.5)/(√(49)))<Z<(10-8.2)/((1.5)/(√(49)))) =P(-14.933< Z<8.4)`

And using a calculator, excel or the normal standard table we have that:

`P(-14.933<Z<8.4)=1-0 = 1`

Part c

`P(\bar X <6)=P(Z<(6-8.2)/((1.5)/(√(49)))=-10.267)`

And using a calculator, excel or the normal standard table we have that:

`P(Z<-10.267)=0`