Question

For air travelers, one of the biggest complaints is of the waiting time between when the airplane taxis away from the terminal until the flight takes off. This waiting time is known to have a right skewed distribution with a mean of 10 minutes and a standard deviation of 8 minutes. Suppose 100 fights have been randomly sampled. Describe the sampling distribution of the mean waiting time between when the airplane taxis away from the terminal until the flight takes off for these 100 flights.

Answer 1

Answer:

The sampling distribution of mean waiting time is $N(10, 0.80^{2})$.

Step-by-step explanation:

Let X = waiting time between when the airplane taxis away from the terminal until the flight takes off.

The distribution of X is right skewed.

The mean of X is, μ = 10 minutes and standard deviation is, σ = 8 minutes.

A sample of 100 flights is selected.

According to the central limit theorem if a large sample (n > 30) is selected randomly from an unknown population with mean μ and standard deviation σ then the sampling distribution of sample mean follows a Normal distribution.

The mean of the sampling distribution of sample mean is:

$\mu_{\bar x}=\mu$

The standard deviation of the sampling distribution of sample mean is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

The sampling distribution of the mean waiting time will also be Normally distributed since the sample size n = 100 > 30.

The mean of the sampling distribution of the mean waiting time is:

$\mu_{\bar x}=\mu=10\ minutes$

The standard deviation of the sampling distribution of the mean waiting time is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{8}{\sqrt{100}}=0.80\ minutes$

Thus, the sampling distribution of mean waiting time is $N(10, 0.80^{2})$.