Normal Distribution Using a larger data set than the one given for Exercises 1-4, assume that airline arrival delays are normally distributed with a mean of -5.0 min and a standard deviation of 30.4 min.

a. Find the probability that a randomly selected flight has an arrival delay time of more than 15 min.

b. Find the value of $Q_3$, the arrival delay time that is the third quartile

It is given that the arrival delay times of flights are normally distributed with a mean equal to-5.0 minutes and a standard deviation equal to 30.4 minutes.

The population mean given is equal to $\mu =-5.0$minutes.

The population standard deviation given is equal to $\sigma =30.4$ minutes.

The sample value given is equal to x=15 minutes.

The z-score for the sample value of 15 minutes is computed below.

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{15-\left(-5.0\right)}{30.4} \\ =0.66 \end{gathered}$$

Thus, z=0.66.

The probabilities corresponding to z-scores are obtained by using the standard normal table.

a.

The probability that a randomly selected flight has an arrival delay time of more than 15 minutes is equal to

$$\begin{gathered} P\left(x>15\right)=P\left(z>0.66\right) \\ =1-P\left(z<0.66\right) \\ =1-0.7454 \\ =0.2546 \end{gathered}$$

Therefore, the probability that a randomly selected flight has an arrival delay time of more than 15 minutes is equal to 0.2546.

b.

It is known that 75% of all the values are less than or equal to the third quartile.

In terms of standard normal probability, the third quartile can be expressed as follows.

$$\begin{gathered} Q_3=P\left(Z⩽z\right) \\ =0.75 \end{gathered}$$

The value of the z-score corresponding to the left-tailed probability of 0.75 is equal to 0.68.

By converting the z-score to the sample value, the following value is obtained:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ x=\mu +z\sigma \\ =-5.0+\left(0.68\right)\left(30.4\right) \\ =15.672 \end{gathered}$$

Thus, the value of the third quartile is equal to 15.672 minutes.