The scheduled arrival time for a daily flight from Boston to New York is 9:30 am. Historical data show that the arrival time follows the continuous uniform distribution with an early arrival time of 9:05 am and a late arrival time of 9:55 am. a. After converting the time data to a minute scale, calculate the mean and the standard deviation of the distribution. (Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) Mean minutes Standard deviation minutes b. What is the probability that a flight arrives late (later than 9:30 am)

Let A the random variable that represent "The arrival time (minutes) for a daily flight from Boston to New York ". And we know that the distribution of A is given by:

[tex]A\sim Uniform(a=0 ,b=50)[/tex]

We select our starting point a=0 representing 9:05 AM and the amount of minutes between 9:05 am to 9:55 am are 50 ans for this reason b =50

For ths uniform distribution the expected value is given by [tex]E(X) =\frac{a+b}{2}[/tex] where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:

[tex]E(A)=\frac{0+50}{2}=25 min[/tex]

And the variance is given by:

[tex]Var(A)= \frac{(b-a)^2}{12}=\frac{(50-0)^2}{12}=208.3333 min^2[/tex]

And the standard deviation by definition is just the square root of the variance:

[tex] Sd(A) =\sqrt{208.3333 min^2}=14.43 min[/tex]

Part b

If we convert 9:30 AM to our scale we know that we have 25 minutes between 9:05 AM and 9:30 AM.

For this case we want to find this probability: [tex] P(A>25)[/tex]

And we can find this probability using the complement rule like this:

[tex] P(A>25) = 1-P(A\leq 25) [/tex]

And we can use the cumulative distribution function given by:

[tex] F(A) = \frac{A-a}{b-a}[/tex]

And we got:

[tex] P(A>25) = 1-P(A\leq 25)=1-\frac{25-0}{50-0}= 1-0.5 = 0.5 [/tex]