Flight times An airline flies the same route at the same time each day. The flight time varies according to a Normal distribution with unknown mean and standard deviation. On $15\%$ of days, the flight takes more than an hour. On $3\%$ of days, the flight lasts $75$ minutes or more. Use this information to determine the mean and standard deviation of the flight time distribution.

The flight takes more than one hour $15\%$ of the time.

On about $3\%$ of days, the flight is $75$ minutes or longer.

On $15\%$of the days, the flight takes longer than an hour.

That means

On $85$percent of days, the flight takes less than an hour.

Now,

In the normal probability table of the appendix, find the $z-$score that corresponds to a probability of $85$percent (or $0.85$)

Note that

The probability that comes closest is $0.8508$which is found in row $1.0$and 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,

In the normal probability table of the appendix, find the $z-$score that corresponds to a probability of $97$percent (or $0.97$)

Note that

The probability that comes closest is $0.9699$which is found in row $1.8$and column$.08$of the normal probability table.

Then

The corresponding $z-$score,

$z=1.88$

Now,

The value will be the mean multiplied by the product of the $z-$value and the (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:

$$\begin{gathered} \mu =60-1.04\sigma \\ =60-1.04(17.8571) \\ =41.4286 \end{gathered}$$