Respuesta :
Answer:
a. k = 3
b. Cumulative distribution function X, [tex]F(x)=\left \{ {0} , x\leq 1 \atop {1-x^{-3}, x>1}} \right.[/tex]
c. Probability when headway exceeds 2 seconds = 0.125
Probability when headway is between 2 and 3 seconds = 0.088
d. Mean value of headway = 1.5
Standard deviation of headway = 0.866
e. Probability that headway is within 1 standard deviation of the mean value = 0.9245
Step-by-step explanation:
From the information provided,
Let X be the time headway between two randomly selected consecutive cars (sec).
The known distribution of time headway is,
[tex]f(x) = \left \{ {\frac{k}{x^4} , x > 1} \atop {0} , x \leq 1 } \right.[/tex]
a. Value of k.
Since the distribution of X is a valid density function, the total area for density function is unity. That is,
[tex]\int\limits^{\infty}_{-\infty} f(x)dx=1[/tex]
So, the equation becomes,
[tex]\int\limits^{1}_{-\infty} f(x)dx + \int\limits^{\infty}_{1} f(x)dx=1\\0 + \int\limits^{\infty}_{1} {\frac{k}{x^4}}.dx=1\\0 + k \int\limits^{\infty}_{1} {\frac{1}{x^4}}.dx=1\\k[\frac{x^{-3}}{-3}]^{\infty}_1=1\\k[0-(\frac{1}{-3})]=1\\\frac{k}{3}=1\\k=3[/tex]
b. For this problem, the cumulative distribution function is defined as :
[tex]F(x) = \int\limits^1_{\infty} f(x)dx + \int\limits^x_1 f(x)dx[/tex]
Now,
[tex]F(x) = 0 + \int\limits^x_1 {\frac{k}{x^4}}.dx\\= 0 + \int\limits^x_1 3x^{-4}.dx\\= 3 \int\limits^x_1 x^{-4}dx\\= 3[\frac{x^{-4+1}}{-4+1}]^3_1\\= 3[\frac{x^{-3}}{-3}]^3_1\\=(\frac{-1}{x^3})|^x_1\\=(-\frac{1}{x^3}-(\frac{-1}{1}))=1- \frac{1}{x^3}=1-x^{-3}[/tex]
Therefore the cumulative distribution function X is,
[tex]F(x)=\left \{ {0} , x\leq 1 \atop {1-x^{-3}, x>1}} \right.[/tex]
c. Probability when the headway exceeds 2 secs.
Using cdf in part b, the required probability is,
[tex]P(X>2)=1-P(X\leq 2)\\=1-F(2)\\=1-[1-2^{-3}]\\=1-(1- \frac{1}{8})\\=\frac{1}{8} = 0.125[/tex]
Probability when headway is between 2 seconds and 3 seconds
Using the cdf in part b, the required probability is,
[tex]P(2<X<3)=P(X<3)-P(X<2)\\=F(3)-F(2)\\=(1-3^{-3})-(1-2^{-3})\\=1-3^{-3}-1+2^{-3}\\=2^{-3}-3^{-3}\\=\frac{1}{8} -\frac{1}{27} \\=0.125-0.03704\\=0.08796[/tex]
≅ 0.088
d. Mean value of headway,
[tex]E(X)=\int\limits x * f(x)dx\\=\int\limits^{\infty}_1 x(3x^{-4})dx\\=3 \int\limits^{\infty}_1 x(x^{-4})dx\\=3 \int\limits^{\infty}_1 x^{-3}dx\\=3[\frac{x^{-3+1}}{-3+1}]^{\infty}_1\\=3[\frac{x^{-2}}{-2}]^{\infty}_1\\=3[\frac{1}{-2x^2}]^{\infty}_1\\=3[- \frac{1}{2x^2}]^{\infty}_1\\=3[- \frac{1}{2(\infty)^2}- (- \frac{1}{2(1)^2})]\\=3(\frac{1}{2})=1.5[/tex]
And,
[tex]E(X^2)= \int\limits^{\infty}_1 x^2(3x^{-4})dx\\=3 \int\limits^{\infty}_1 x^{-2} dx\\=3[- \frac{1}{x}]^{\infty}_1\\=3(- \frac{1}{\infty}+1)=3[/tex]
The standard deviation of headway is,
[tex]= \sqrt{V(X)}\\ =\sqrt{E(X^2)-[E(X)]^2} \\=\sqrt{3-(1.5)^2} \\=0.8660254[/tex]
≅ 0.866
e. Probability that headway is within 1 standard deviation of the mean value
[tex]P(\alpha - \beta < X < \alpha + \beta) = P(1.5-0.866 < X < 1.5 +0.866)\\=P(0.634 < X < 2.366)\\=P(X<2.366)-P(X<0.634)\\=F(2.366)-F(0.634)\\=F(2.366)-0[/tex]
From part b, F(x) = 0, if x ≤ 1
[tex]=1-(2.366)^{-3}\\=0.9245[/tex]
