Question

(10 points) Suppose a flight is about to land and the announ land is 30 minutes with a standard deviation of 8.66 minutes. The flights are uniformly distributed. Find the appropriate interval of time such that its distribution is a probability density function me to 4.

Answer 1

Given that

Mean = 30

Standard deviation = 8.66

Here,

uniform distribution mean and variance with parameters 'a' and 'b'

Mean = (a + b) / 2

Variance = (b - a)² / 12

Calculations

Mean => 30 = (b + a) / 2

b + a = 60...........(1)

Variance => (8.66)² = (b - a)² / 12

(b - a)² = 900 (rounded as integer)

b - a = 30............(2)

By solving (1) and (2), we have

b = 45

a = 15

Therefore required interval (15, 45)

And probability density function, f(x) = 1 / (45 - 15),