Chapter 2.2, Problem 61E

To determine

the mean and standard deviation of the flight time distribution

Answer to Problem 61E

The required mean and standard deviation are 17.86min and 41.43min respectively.

Explanation of Solution

Given:

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

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

Calculation:

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

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

Similarly, P (flight lasts 75 minutes or more) = 0.03

Then, 97% of flights last less than 75 minutes, he z value corresponding to 0.97 probability from standard normal table is 1.88.

The following graph illustrates the given information to set up two equations in two unknowns.

  

Then, we can find the values of unknown mean and standard deviation using the following two equations as follows:

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

Multiplying both sides of the equations by and subtracting yields

  $0.84\sigma =15$

or $\boxed{\sigma =17.86\text{minutes}}$

Substituting this value back into the equation (1), we obtain

  $1.04=\frac{60-\mu }{17.86}$

or $\mu =60-1.04\left(17.86\right)$

  $\boxed{41.43\text{minutes}}$

Hence, the required mean and standard deviation are 17.86min and 41.43min respectively.

Conclusion:

Hence, the required mean and standard deviation are 17.86min and 41.43min respectively.