Question

A manufacturing company has developed a fuel-efficient machine that combines pressure washing with steam cleaning. It is designed to deliver 7.0 gallons of cleaner per minute at 1,000 pounds per square inch of pressure washing. In fact, it delivers an amount at random anywhere between 5.5 and 8.5 gallons per minute. Assume that the RV X, the amount of cleaner delivered, is an uniform RV with probability density f(x) = 1/2 for 6.5 < x < 8.5. What is the probability that more than 7.3 gallons of cleaner are dispensed per minute?

Answer 1

Answer:

$P(X>7.3) = \int_{5.5}^{7.3} \frac{1}{2} dx = \frac{1}{2} x \Big|_{6.5}^{7.3} = \frac{1}{2} (7.3-6.5)= 0.4$

And then we can find the probability replacing the last result and we got:

$P(X>7.3)= 1- P(X$

Step-by-step explanation:

For this case we have the following density function given:

$f(x) = \frac{1}{2} , 5.5 \leq X \leq 8.5$

And we want to find the following probability:

$P(X>7.3)$

We can find this probability with the complement rule like this:

$P(X>7.3)= 1- P(X$

And we can find the following probability:

$P(X>7.3) = \int_{5.5}^{7.3} \frac{1}{2} dx = \frac{1}{2} x \Big|_{6.5}^{7.3} = \frac{1}{2} (7.3-6.5)= 0.4$

And then we can find the probability replacing the last result and we got:

$P(X>7.3)= 1- P(X$