Respuesta :
Answer:
(a) The probability that the distance between two flaws is greater than 15 m is 0.2865.
(b) The probability that the distance between two flaws is between 8 and 20 m is 0.3246.
(c) The median is 8.322.
(d) The standard deviation is 12.
(e) The 65th percentile of the distances is 12.61 m.
Step-by-step explanation:
The random variable X can be defined as the distance between flaws on a long cable.
The random variable X is exponentially distributed with mean, μ = 12 m.
The parameter of the exponential distribution is:
[tex]\lambda=\frac{1}{\mu}=\frac{1}{12}=0.0833[/tex]
The probability density function of X is:
[tex]f_{X}(x)=0.0833e^{-0.0833x};\ x\geq 0[/tex]
(a)
Compute the probability that the distance between two flaws is greater than 15 m as follows:
[tex]P(X\geq15)=\int\limits^{\infty}_{15}{0.0833e^{-0.0833x}}\, dx\\=0.0833\times \int\limits^{\infty}_{15}{e^{-0.0833x}}\, dx\\=0.0833\times |\frac{e^{-0.0833x}}{-0.0833}|^{\infty}_{15}\\=e^{0.0833\times 15}\\=0.2865[/tex]
Thus, the probability that the distance between two flaws is greater than 15 m is 0.2865.
(b)
Compute the probability that the distance between two flaws is between 8 and 20 m as follows:
[tex]P(8\leq X\leq20)=\int\limits^{20}_{8}{0.0833e^{-0.0833x}}\, dx\\=0.0833\times \int\limits^{20}_{8}{e^{-0.0833x}}\, dx\\=0.0833\times |\frac{e^{-0.0833x}}{-0.0833}|^{20}_{8}\\=e^{0.0833\times 8}-e^{0.0833\times 20}\\=0.51355-0.1890\\=0.32455\\\approx0.3246[/tex]
Thus, the probability that the distance between two flaws is between 8 and 20 m is 0.3246.
(c)
The median of an Exponential distribution is given by:
[tex]Median=\frac{\ln (2)}{\lambda}[/tex]
Compute the median as follows:
[tex]Median=\frac{\ln (2)}{\lambda}[/tex]
[tex]=\farc{0.69315}{0.08333}\\=8.322[/tex]
Thus, the median is 8.322.
(d)
The standard deviation of an Exponential distribution is given by:
[tex]\sigma=\sqrt{\frac{1}{\lambda^{2}}}[/tex]
Compute the standard deviation as follows:
[tex]\sigma=\sqrt{\frac{1}{\lambda^{2}}}[/tex]
[tex]=\sqrt{\frac{1}{0.0833^{2}}}\\=12.0048\\\approx 12[/tex]
Thus, the standard deviation is 12.
(e)
Let x be 65th percentile of the distances.
Then, P (X < x) = 0.65.
Compute the value of x as follows:
[tex]\int\limits^{x}_{0}{0.0833e^{-0.0833x}}\, dx=0.65\\0.0833\times \int\limits^{x}_{0}{e^{-0.0833x}}\, dx=0.65\\0.0833\times |\frac{e^{-0.0833x}}{-0.0833}|^{x}_{0}=0.65\\-e^{-0.0833x}+1=0.65\\-e^{-0.0833x}=-0.35\\-0.0833x=-1.05\\x=12.61[/tex]
Thus, the 65th percentile of the distances is 12.61 m.
