Final answer:
To determine the value of k, we need to find the value for which f(x) is a legitimate pdf. The value of k is 2. The cumulative distribution function is F(x) = { k * x for x ≤ 1; k/2 * x^2 for x > 1 }. The probability that the headway exceeds 2 seconds is 4, and the probability that the headway is between 2 and 3 seconds is 2. The mean value of the headway is 1 second, and the standard deviation is approximately 1.040 seconds. The probability that the headway is within 1 standard deviation of the mean value is approximately 2.079.
Step-by-step explanation:
To determine the value of k for which f(x) is a legitimate pdf, we need to find the value that makes the integral of f(x) over its entire domain equal to 1.
Since f(x) is defined differently for x values greater than 1 and less than or equal to 1, we need to integrate those two cases separately.
Integrating f(x) for x > 1:
∫f(x)dx = ∫kx dx = k/2 * x^2 | 1 to ∞
Integrating f(x) for x ≤ 1:
∫f(x)dx = ∫k dx = k * x | 0 to 1
Setting the two integrals equal to 1 and solving for k gives us:
k/2 * ∞^2 - k/2 * 1^2 = 1
k * 1 - k * 0 = 1
Simplifying and solving for k, we get:
k = 2
Therefore, the value of k for which f(x) is a legitimate pdf is 2.
To obtain the cumulative distribution function (CDF), we integrate f(x) from 0 to x for x ≤ 1, and integrate from 1 to x for x > 1.
For x ≤ 1:
F(x) = ∫f(x)dx = ∫k dx = k * x | 0 to x
For x > 1:
F(x) = ∫f(x)dx = ∫kx dx = k/2 * x^2 | 1 to x
Therefore, the cumulative distribution function is:
F(x) = { k * x for x ≤ 1; k/2 * x^2 for x > 1 }
To determine the probability that the headway exceeds 2 seconds, we need to find the value of F(2). Plugging in 2 into the CDF equation, we get:
F(2) = 2 * 2 = 4
Therefore, the probability that the headway exceeds 2 seconds is 4.
To determine the probability that the headway is between 2 and 3 seconds, we need to find the difference between F(3) and F(2). Plugging in 3 into the CDF equation, we get:
F(3) = 2 * 3 = 6
The probability that the headway is between 2 and 3 seconds is the difference between F(3) and F(2):
F(3) - F(2) = 6 - 4 = 2
Therefore, the probability that the headway is between 2 and 3 seconds is 2.
To obtain the mean value of the headway, we integrate x * f(x) over its entire domain. For x > 1, the mean is given by:
∫xf(x)dx = ∫kx^2 dx = k/3 * x^3 | 1 to ∞
For x ≤ 1, the mean is given by:
∫xf(x)dx = ∫kx dx = k/2 * x^2 | 0 to 1
Adding the two results together and solving for the mean, we get:
Mean = k/3 * ∞^3 - k/3 * 1^3 + k/2 * 1^2 = ∞ - 1/3 + 1/2
Since the value of k is 2, we can substitute it into the equation:
Mean = 2/3 - 1/3 + 1/2 = 1
Therefore, the mean value of the headway is 1 second.
To obtain the standard deviation of the headway, we need to calculate the variance first. The variance is given by:
Var = ∫(x - mean)^2 * f(x)dx
Using the same integration bounds as the mean calculation, we get:
Var = ∫((x - mean)^2) * f(x)dx = ∫((x - 1)^2) * f(x)dx
Solving this integral, we get:
Var = 1.083
Therefore, the standard deviation of the headway is approximately 1.040 seconds.
To calculate the probability that the headway is within 1 standard deviation of the mean value, we need to find the values of F(mean - std) and F(mean + std). Plugging in the mean and standard deviation values into the CDF equation, we get:
F(mean - std) = 2 * (mean - std) for mean - std ≤ 1; k/2 * (mean - std)^2 for mean - std > 1
F(mean + std) = 2 * (mean + std) for mean + std ≤ 1; k/2 * (mean + std)^2 for mean + std > 1
Substituting the mean and standard deviation values, we get:
F(mean - std) = 2 * (1 - 1.040) for 1 - 1.040 ≤ 1; 2/3 * (1 - 1.040)^2 for 1 - 1.040 > 1
F(mean + std) = 2 * (1 + 1.040) for 1 + 1.040 ≤ 1; 2/3 * (1 + 1.040)^2 for 1 + 1.040 > 1
Calculating these values, we get:
F(mean - std) = 0
F(mean + std) = 2.079
Therefore, the probability that the headway is within 1 standard deviation of the mean value is 2.079 minus 0, which is approximately 2.079.