Question

The scheduled arrival time for a daily flight from Boston to New York is 9:35 am. Historical data show that the arrival time follows the continuous uniform distribution with an early arrival time of 9:18 am and a late arrival time of 9:41 am.
After converting the time data to a minute scale, calculate the mean and the standard deviation for the distribution.

Answer 1

Answer:

The mean is 11.5 minutes and the standard deviation is of 6.64 minutes

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The mean is:

$M = \frac{a+b}{2}$

The standard deviation is:

$S = \sqrt{\frac{(b-a)^{2}}{12}}$

Arrival time of 9:18 am and a late arrival time of 9:41 am.

9:41 is 23 minutes from 9:18. So the time is uniformily distributed between 0 and 23 minutes, so a = 0, b = 23.

Mean:

$M = \frac{a+b}{2} = \frac{0+23}{2} = 11.5$

Standard deviation:

$S = \sqrt{\frac{(b-a)^{2}}{12}} = \sqrt{\frac{(23 - 0)^{2}}{12}} = 6.64$

The mean is 11.5 minutes and the standard deviation is of 6.64 minutes