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 9:07 am and a late arrival time of 9:57 am. a. After converting the time data to a minute scale, calculate the mean and the standard deviation for the distribution. (Round your answers to 2 decimal places.)Mean:minutes Standard deviation: minutes b. What is the probability that a flight arrives late (later than 9:30 am)? (Do not round intermediate calculations. Round your answer to 2 decimal places.)Probability:

Step-by-step explanation:

Let x be a random variable representing the flight arrival time from Boston to New York.

For a uniform probability distribution, the notation is

X U(a, b) where a is the lowest value of x and b is the lowest value of x

The probability density function, f(x) = 1/(b - a)

Mean, µ = (a + b)/2

Standard deviation, σ = √(b - a)²/12

From the information given, the time difference in minutes is 9:57 - 9:07 = 50 minutes. Therefore,

a = 0

b = 50

µ = (0 + 50)/2 = 25

σ = √(50 - 0)²/12 = 14.43

b) converting to minutes, it is 9:30 - 9:07 = 23 minutes

the probability that a flight arrives late(later than 9:30 am) is expressed as P(x > 23)

f(x) = 1/(50) = 0.02

P(x > 23) = (50 - 23)0.02 = 0.54

(0.2)+(-8)=0.3+-3+0.3m is the equation i guess? ?

step-by-step explanation:

never fear papaguy is here

answer