Question

The warranty on a machine specifies that it will be replaced at failure or age 4, whichever occurs first. The machines age at failure, X, has density function:
f(x) = 1/5 for 0 0 otherwise

Let Y be the age of the machine at the time of replacement.
Determine the variance of Y.

Answer 1

Answer:

$Var(Y) = E(Y^2) -[E(Y)]^2 = \frac{112}{5} -[\frac{12}{5}]^2 =\frac{128}{75}$

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

Solution to the problem

Since the warranty on a machine specifies that it will be replaced at failure or age 4 and the distribution for X is defined between 0 and 5 then if we define the random variable Y ="the age of the machine at the time of replacement" we know that the values for Y needs to be between 0 and 4 or between 4 and and we can define the following density function:

$f(y) = x, 0 \leq x \leq 4$

$f(y) = 4, 4 \leq x \leq 5$

$f(y) = 0$ for other case

Now we can apply the definition of expected value and we have this:

$E(Y) = \int_{0}^4 \frac{1}{5} x dx +\int_{4}^5 \frac{4}{5}dx$

$E(Y) = \frac{1}{10} (4^2-0^2) +\frac{4}{5} (5-4) = \frac{16}{10}+ \frac{4}{5}=\frac{12}{5}$

And for the second moment we have:

$E(Y) = \int_{0}^4 \frac{1}{5} x^2 dx +\int_{4}^5 \frac{16}{5}dx$

$E(Y^2) = \frac{1}{15} (4^3-0^3) +\frac{16}{5} (5-4) = \frac{64}{15}+ \frac{16}{5}=\frac{112}{5}$

And the variance would be given by:

$Var(Y) = E(Y^2) -[E(Y)]^2 = \frac{112}{5} -[\frac{12}{5}]^2 =\frac{128}{75}$