.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 localid="1649758011998" $3\%$of days, the flight lasts localid="1649758017692" $75$minutes or more. Use this information to determine the mean and standard deviation of the flight time distribution

Given P (flight takes more than an hour )$=0.15$

If $15$of days, all fights take more than an hour, then in remaining $85\%$of days, fights will take less than an hour. The $z$value corresponding to $0.85$probability from standard normal table is $1.04$

Similarly. P (fight lasts $75$minutes of more) $=0.03$

Then, $97\%$of flights lasts less than $75$minutes, he $z$value corresponding to $0.97$probability from standard nomal table is$1.86$

The graph below shows how to solve two equations involving two unknowns, given the given information.

Following these two equations, we can calculate the unknown mean and standard deviation:

$$\begin{gathered} 1.04=\frac{60-\mu }{\sigma }----\left(1\right) \\ 1.88=\frac{75-\mu }{\sigma }----\left(2\right) \end{gathered}$$

A multiplication and subtraction of both sides of the equation yields

localid="1649998199753" $$\begin{gathered} 0.84\sigma =15 \\ or\sigma =17.86 \end{gathered}$$

We obtain the following when we substitute this value back into equation (1):

localid="1649998214444" $$\begin{gathered} 1.04=\frac{60-\mu }{17.86} \\ or\mu =60-1.04(17.86) \\ =41.43minutes \end{gathered}$$

As a result, the required mean and standard deviation are$17.86minand41.43min$respectively.