91Ó°ÊÓ

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:

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

A multiplication and subtraction of both sides of the equation yields

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

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

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

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