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

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:

[tex]X \sim N(8.2,1.5)[/tex]

Where [tex]\mu=8.2[/tex] and [tex]\sigma=1.5[/tex]

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

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

We can find the individual probabilities like this:

Part a

[tex]P(\bar X <10)=P(Z<\frac{10-8.2}{\frac{1.5}{\sqrt{49}}}=8.4)[/tex]

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

[tex]P(Z<8.4)=1[/tex]

Part b

[tex]P(5<\bar X <10)=P(\frac{5-8.2}{\frac{1.5}{\sqrt{49}}}<Z<\frac{10-8.2}{\frac{1.5}{\sqrt{49}}}) =P(-14.933< Z<8.4)[/tex]

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

[tex]P(-14.933<Z<8.4)=1-0 = 1[/tex]

Part c

[tex]P(\bar X <6)=P(Z<\frac{6-8.2}{\frac{1.5}{\sqrt{49}}}=-10.267)[/tex]

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

[tex]P(Z<-10.267)=0[/tex]