The time until failure of an electronic device has an

Chapter 6, Problem 100SE

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QUESTION:

The time until failure of an electronic device has an exponential distribution with mean 15 months. If a random sample of five such devices are tested, what is the probability that the first failure among the five devices occurs

a after 9 months?

b before 12 months?

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QUESTION:

The time until failure of an electronic device has an exponential distribution with mean 15 months. If a random sample of five such devices are tested, what is the probability that the first failure among the five devices occurs

a after 9 months?

b before 12 months?

ANSWER:

Step 1 of 3

Let \(X_{1}, X_{2}, \ldots X_{5}\) are independent exponential distribution with a mean of 15 months.

The probability density function of \(X_{i}\) is

\(f_{x}(x)=\left\{\begin{array}{ll}
\frac{1}{15} e^{-\frac{x}{15}} & ; x>0 \\
0 & ; \text { Otherwise }
\end{array}\right.\)

The CDF of the exponential distribution is

\(F_{X}(x)=1-e^{-\frac{x}{15}}\)

Let Consider \(X_{(1)}=\min \left\{x_{1}, x_{2}, \ldots, x_{5}\right\}\) denote the first failure length among the five devices.

Now the pdf of \(X_{(1)}\) is

\(\begin{aligned}
f_{x_{(1)}}(x) & =n[1-F(x)]^{n-1} f(x) \\
& =5\left[1-\left(1-e^{-\frac{x}{15}}\right)\right]^{5-1} \frac{1}{15} e^{-\frac{x}{15}} \\
& =\frac{5}{15} e^{-\frac{4 x}{15}} e^{-\frac{x}{15}} \\
& =\frac{1}{3} e^{-\frac{x}{3}}
\end{aligned}\)

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