Question

Suppose that an airline quotes a flight time of 128 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 102 and 154 minutes. Letting the time unit be one minute, (a) Write the formula for the probability curve of x. b)P(129 LESS THAN OR EQUAL TOO X LESS THAN OR EQUAL TO 146 C)Find the probability that a randomly selected flight between the two cities will be at least 3 minutes late. ROUND ALL ANSWERS 4 DECIMAL PLACES

Answer 1

Answer:

(a) The probability density function of X is:

$f_{X}(x)=\frac{1}{b-a};\ a$

(b) The value of P (129 ≤ X ≤ 146) is 0.3462.

(c) The probability that a randomly selected flight between the two cities will be at least 3 minutes late is 0.4327.

Step-by-step explanation:

The random variable X is defined as the flight time between the two cities.

Since the random variable X denotes time interval, the random variable X is continuous.

(a)

The random variable X is Uniformly distributed with parameters a = 10 minutes and b = 154 minutes.

The probability density function of X is:

$f_{X}(x)=\frac{1}{b-a};\ a$

(b)

Compute the value of P (129 ≤ X ≤ 146) as follows:

Apply continuity correction:

P (129 ≤ X ≤ 146) = P (129 - 0.50 < X < 146 + 0.50)

                           = P (128.50 < X < 146.50)

                           $=\int\limits^{146.50}_{128.50} {\frac{1}{154-102}} \, dx$

                           $=\frac{1}{52}\times \int\limits^{146.50}_{128.50} {1} \, dx$

                           $=\frac{1}{52}\times (146.50-128.50)$

                           $=0.3462$

Thus, the value of P (129 ≤ X ≤ 146) is 0.3462.

(c)

It is provided that a randomly selected flight between the two cities will be at least 3 minutes late, i.e. X ≥ 128 + 3 = 131.

Compute the value of P (X ≥ 131) as follows:

Apply continuity correction:

P (X ≥ 131) = P (X > 131 + 0.50)

                 = P (X > 131.50)

                 $=\int\limits^{154}_{131.50} {\frac{1}{154-102}} \, dx$

                 $=\frac{1}{52}\times \int\limits^{154}_{131.50} {1} \, dx$

                 $=\frac{1}{52}\times (154-131.50)$

                 $=0.4327$

Thus, the probability that a randomly selected flight between the two cities will be at least 3 minutes late is 0.4327.