Respuesta :
Answer:
[tex] Var(Y) = E(Y^2) -[E(Y)]^2 = \frac{112}{5} -[\frac{12}{5}]^2 =\frac{128}{75}[/tex]
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:
[tex]f(y) = x, 0 \leq x \leq 4[/tex]
[tex] f(y) = 4, 4 \leq x \leq 5[/tex]
[tex] f(y) = 0[/tex] for other case
Now we can apply the definition of expected value and we have this:
[tex] E(Y) = \int_{0}^4 \frac{1}{5} x dx +\int_{4}^5 \frac{4}{5}dx[/tex]
[tex] E(Y) = \frac{1}{10} (4^2-0^2) +\frac{4}{5} (5-4) = \frac{16}{10}+ \frac{4}{5}=\frac{12}{5}[/tex]
And for the second moment we have:
[tex] E(Y) = \int_{0}^4 \frac{1}{5} x^2 dx +\int_{4}^5 \frac{16}{5}dx[/tex]
[tex] E(Y^2) = \frac{1}{15} (4^3-0^3) +\frac{16}{5} (5-4) = \frac{64}{15}+ \frac{16}{5}=\frac{112}{5}[/tex]
And the variance would be given by:
[tex] Var(Y) = E(Y^2) -[E(Y)]^2 = \frac{112}{5} -[\frac{12}{5}]^2 =\frac{128}{75}[/tex]
