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)

a)

b)

Step-by-step explanation:

Part a

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:

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 where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:

And the variance is given by:

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

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:

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

And we can use the cumulative distribution function given by:

And we got:

true statements:

a) x + y = y + x   ⇒ commutative property of addition

b) (x * y) * z = x * (y * z) ⇒ associative property of multiplication

d) (x + y) + z = x + (y + z) ⇒ associative property of addition

false statements:

c) x - y = y - x   should be: x - y = - y + x

e) (x - y) - z = x - (y - z) should be: (x - y) - z = x - (y + z)

step-by-step explanation: