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:
[tex]M = \frac{a+b}{2}[/tex]
The standard deviation is:
[tex]S = \sqrt{\frac{(b-a)^{2}}{12}}[/tex]
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:
[tex]M = \frac{a+b}{2} = \frac{0+23}{2} = 11.5[/tex]
Standard deviation:
[tex]S = \sqrt{\frac{(b-a)^{2}}{12}} = \sqrt{\frac{(23 - 0)^{2}}{12}} = 6.64[/tex]
The mean is 11.5 minutes and the standard deviation is of 6.64 minutes
