The probability density function of X , the lifetime of a certain type of electronic device (measured in hours), is given by f ( x ) = { 10 x 2 x > 10 0 x ≤ 10 Find P { X > 20 } . What is the cumulative distribution function of X ? What is the probability that of 6 such types of devices, at least 3 will function for at least 15 hours? What assumptions are you making?

The cumulative distribution function of a random variable accumulates the probability from left side up to a point. The solutions are:

The cumulative distribution function of X is: [tex]F(x) = 1 - 10/x[/tex] ( for x > 10)
The probability that of 6 such types of devices, at least 3 will function for at least 15 hours is 0.33 approx

What is cumulative distribution function?

Suppose that for a random variable X, its probability function be f(x).

Then we have:

[tex]CDF = F(x) = f(X \leq x)[/tex]

If probability function is integrable, then:

[tex]F(X) = \int_{-\infty}^xf(t)dt[/tex]

For the considered case, let we have:

X = lifetime of a certain type of electronic device in hours, then the probability function of X is defined as:

[tex]f(x) = \left \{\large {{10/x^2, x > 10} \atop \large {0,\: x\leq 10}} \right.[/tex]

Therefore, the cumulative distribution function of X is:

[tex]F(X) = \int_{-\infty}^xf(t)dt = \int_{-\infty}^{10} 0dt + \int_{10}^x10/(t)^2dt\\\\F(X) = [\dfrac{-10}{t}]{{t =x} \atop {t=10}} = 1-\dfrac{10}{x}[/tex](for all x > 10)

Then, the probability that one such device will work at least 15 hours is:

[tex]P(X \geq 15) = 1 - P(X < 15) = 1 - F(15) = 1 -\dfrac{10 }{15} \approx 0.33[/tex]

Learn more about cumulative distribution functions here:

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