Chapter 2.2, Problem 71E

To determine

Mean and standard deviation of the flight time distribution.

Answer to Problem 71E

Mean,

  $\mu =41.4286$

Standard deviation,

  $\sigma =17.8571$

Explanation of Solution

Given information:

On 15% of days, the flight takes more than an hour.

On 3% of days, the flight lasts 75 minutes or more.

Calculations:

The flight takes more than one hour on 15% of days.

That means

The flight takes less than one hour on 85% of days.

Now,

Find the z − score corresponding the probability of 85% (or 0.85) in the normal probability table of the appendix.

Note that

The closest probability would be 0.8508 that lies in the row 1.0 and in the column .04 of the normal probability table.

Then

The corresponding z − score,

  $z=1.04$

And

The flight takes more than 75 minutes on 3% of days.

That means

The flight takes less than 75 minutes on 97% of days.

Now,

Find the z − score corresponding the probability of 97% (or 0.97) in the normal probability table of the appendix.

Note that

The closest probability would be 0.9699 that lies in the row 1.8 and in the column .08 of the normal probability table.

Then

The corresponding z − score,

  $z=1.88$

Now,

The value will be the mean increased by the product of the z − value and the $\sigma$ (standard deviation).

  $60=x=\mu +z\sigma =\mu +1.04\sigma$

And

  $75=x=\mu +z\sigma =\mu +1.88\sigma$

Subtract the above two equations:

  $15=0.84\sigma$

Divide both sides by 0.84:

We have

Standard deviation,

  $\sigma =\frac{15}{0.84}=17.8571$

Then

Find the mean:

  $\begin{array}{l}\mu =60-1.04\sigma \\ =60-1.04\left(17.8571\right)\\ =41.4286\end{array}$