# The probability density function of x the lifetime of a certain type of electronic device

The probability density function f(x) of a random variable x is given below, where x is the lifetime of a certain type of electronic device (measured in hours):

$f(x) =\Bigg\{\begin{array}{rr} \dfrac{10}{x^2} & x>10\\ 0 & x\leq 10 \\ \end{array}$

– Find the cumulative distribution function $F(x)$ of $x$.

– Find the probability that ${x>20}$.

– Find the probability that out of 6 such types of devices, at least 3 will function for at least 15 hours.

The aim of the question is to cumulative distribution function given a probability density function using the basic concepts of probability theory, calculus, and binomial random variables.

Part (a)

The cumulative distribution function $F(x)$ can be calculated simply by integrating  the probability density function $f(x)$ over $-\infty$ to $+\infty$.

For $x\leq10$,

$F(x) = P(X\leq x) = \int_{-\infty}^{10} f(u) du= 0$

For $x>10$,

$F(x) = P(X\leq x) = \int_{10}^{x} f(u) du= \int_{10}^{x} \frac{10}{x^2} du = 10 \int_{10}^{x} x^{-2} du$

$=10 |(-2+1) x^{-2+1}|_{10}^{x} = 10 |(-1) x^{-1}|_{10}^{x} = -10 |\frac{1}{ x}|_{10}^{x}$

$= -10 (\frac{1}{x}-\frac{1}{10}) = 1-\frac{10}{x}$

Hence,

$F(x) =\Bigg\{\begin{array}{rr} 1-\frac{10}{x} & x>10\\ 0 & x\leq 10 \\ \end{array}$

Part (b)

Since $F(x) = P(X\leq x)$ and $P(x>a) = 1 – P(x \leq a)$,

$P(x>20) = 1 – P(x \leq 20) = 1 – F(20) = 1 – \bigg\{1-\frac{10}{20}\bigg\} = 1 – 1 + \frac{1}{2} = \frac{1}{20}$

Part (c)

To solve this part, we first need to find the probability that a device will operate for at least 15 years i.e. $P(x \leq 15)$. Lets call this probability of success $q$

$q = P(x \leq 15) = F(15) = 1-\frac{10}{15} = \frac{15 – 10}{15} = \frac{5}{15} = \frac{1}{3}$

Consequently, probability of failure $p$ is given by,

$p = 1 – q = 1 – frac{1}{3} = \frac{2}{3}$

The probability of success of k devices out of N can be approximated with a binomial random variable as follows:

$f_K(k) = \binom{N}{k} p^k q^{N-k}$

By using the above formula, we can find the following probabilities:

$\text{Probability of failure of 0 devices out of 6} = f_K(0) = \binom{6}{0} \bigg\{\frac{2}{3}\bigg\}^0 \bigg\{\frac{1}{3}\bigg\}^6 = \frac{1}{729}$

$\text{Probability of failure of 1 devices out of 6} = f_K(1) = \binom{6}{1} \bigg\{\frac{2}{3}\bigg\}^1 \bigg\{\frac{1}{3}\bigg\}^5 = \frac{4}{243}$

$\text{Probability of failure of 2 devices out of 6} = f_K(2) = \binom{6}{2} \bigg\{\frac{2}{3}\bigg\}^2 \bigg\{\frac{1}{3}\bigg\}^4 = \frac{20}{243}$

$\text{Probability of failure of 3 devices out of 6} = f_K(3) = \binom{6}{3} \bigg\{\frac{2}{3}\bigg\}^3 \bigg\{\frac{1}{3}\bigg\}^3 = \frac{160}{729}$

## Numerical Result

$\text{Probability of success of at least 3 devices} = 1 – f_K(0) – f_K(1) – f_K(2) -f_K(3)$

$= 1 – \frac{1}{729} -\frac{4}{243}- \frac{20}{243}-\frac{160}{729} = \frac{496}{729} = 0.68$

## Example

In the same question given above, find the probability that a device will work for at least 30 years.

$P(x \leq 30) = F(30) = 1-\frac{10}{30} = \frac{30 – 10}{30} = \frac{20}{30} = \frac{2}{3}$