Answer :
Answer:
(a) The probability density function of X is:
[tex]f_{X}(x)=\frac{1}{b-a};\ a<X<b,\ a<b[/tex]
(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:
[tex]f_{X}(x)=\frac{1}{b-a};\ a<X<b,\ a<b[/tex]
(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)
[tex]=\int\limits^{146.50}_{128.50} {\frac{1}{154-102}} \, dx[/tex]
[tex]=\frac{1}{52}\times \int\limits^{146.50}_{128.50} {1} \, dx[/tex]
[tex]=\frac{1}{52}\times (146.50-128.50)[/tex]
[tex]=0.3462[/tex]
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)
[tex]=\int\limits^{154}_{131.50} {\frac{1}{154-102}} \, dx[/tex]
[tex]=\frac{1}{52}\times \int\limits^{154}_{131.50} {1} \, dx[/tex]
[tex]=\frac{1}{52}\times (154-131.50)[/tex]
[tex]=0.4327[/tex]
Thus, the probability that a randomly selected flight between the two cities will be at least 3 minutes late is 0.4327.