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pashok25 [27]
3 years ago
15

The distance between flaws on a long cable is exponentially distributed with mean 12 m.

Mathematics
1 answer:
Elden [556K]3 years ago
4 0

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 <em>X</em> can be defined as the distance between flaws on a long cable.

The random variable <em>X</em> is exponentially distributed with mean, <em>μ</em> = 12 m.

The parameter of the exponential distribution is:

\lambda=\frac{1}{\mu}=\frac{1}{12}=0.0833

The probability density function of <em>X</em> is:

f_{X}(x)=0.0833e^{-0.0833x};\ x\geq 0

(a)

Compute the  probability that the distance between two flaws is greater than 15 m as follows:

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

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:

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

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:

Median=\frac{\ln (2)}{\lambda}

Compute the median as follows:

Median=\frac{\ln (2)}{\lambda}

             =\farc{0.69315}{0.08333}\\=8.322

Thus, the median is 8.322.

(d)

The standard deviation of an Exponential distribution is given by:

\sigma=\sqrt{\frac{1}{\lambda^{2}}}

Compute the standard deviation as follows:

\sigma=\sqrt{\frac{1}{\lambda^{2}}}

   =\sqrt{\frac{1}{0.0833^{2}}}\\=12.0048\\\approx 12

Thus, the standard deviation is 12.

(e)

Let <em>x</em> be 65th percentile of the distances.

Then, P (X < x) = 0.65.

Compute the value of <em>x</em> as follows:

\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

Thus, the 65th percentile of the distances is 12.61 m.

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