Suppose that an airline quotes a flight time of 2 hours, 10 minutes between two cities. Furthermore, suppose that historical flight records indicate that the actual flight time between the two cities, x, is uniformly distributed between 2 hours and 2 hours, 20 minutes. Let the time unit be one minute.a. Write the formula for the probability curve of x.b. Graph the probability curve of x.c. Find P(125 < x < 135).

the probability of the actual flight time being between 125 and 135 minutes is 1/2.

a. The range of possible values of x is between 2 hours (i.e., 120 minutes) and 2 hours and 20 minutes (i.e., 140 minutes). Since the distribution is uniform, the probability density function is a constant value over this range, and zero outside of it. Let the probability density function be denoted as f(x), then:

f(x) = 1/(140-120) = 1/20, for 120 ≤ x ≤ 140

f(x) = 0, otherwise

b. To graph the probability density function, we plot f(x) against x for the interval 120 ≤ x ≤ 140, and set f(x) to 0 outside this interval. The graph of the probability density function is a horizontal line segment of height 1/20 over the interval [120, 140], as shown below:

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 120    125                       140

c. We want to find P(125 < x < 135). Since the probability density function is a constant value of 1/20 over the interval [120, 140], the probability of x being between 125 and 135 minutes can be found by finding the area under the probability density function curve between 125 and 135. This area can be computed as follows:

P(125 < x < 135) = ∫125^135 f(x) dx

= ∫125^135 (1/20) dx

= (1/20) [x]125^135

= (1/20) (135 - 125)

= 1/2

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