Question

A machine is set to pump cleanser into a process at the rate of 8 gallons per minute. upon inspection, it is learned that the machine actually pumps cleanser at a rate described by the uniform distribution over the interval 7.5 to 10.5 gallons per minute. find the probability that between 8.0 gallons and 9.0 gallons are pumped during a randomly selected minute.

Answer 1

The probability that between 8.0 gallons and 9.0 gallons are pumped during a randomly selected minute is 0.3749 or 37.49%

The Mean of the uniform distribution for an interval from a to b is given by the formula :

μ = $\frac{a+b}{2}$

Here, the given interval is from 7.5 to 10.5 gallons per minute, that means a= 7.5 and b= 10.5

So, Mean(μ) = $\frac{7.5+10.5}{2}= \frac{18}{2}= 9$

Now, the standard deviation of the uniform distribution is given by the formula:

σ = $\frac{b-a}{\sqrt{12}}$

So, the standard deviation(σ) = $\frac{10.5-7.5}{\sqrt{12}}= \frac{3}{\sqrt{12}}= 0.8660254$

As we need to find the probability that between 8.0 gallons and 9.0 gallons are pumped, so now we will find z-score for both 8.0 and 9.0

The formula for z-score: z = (X - μ)/σ

So, z(X= 8.0) = $\frac{8.0 - 9}{0.8660254} =\frac{-1}{0.8660254}= -1.1547$

and z(X= 9.0) = $\frac{9.0 - 9}{0.8660254} = 0$

Now according to the z-score table, P(z=-1.1547) = 0.1251 and P(z=0) = 0.5

So, P(8.0 < x < 9.0)

= P(-1.1547 < z <0)

= 0.5 - 0.1251

= 0.3749

= 37.49%

Thus, the probability that between 8.0 gallons and 9.0 gallons are pumped during a randomly selected minute is 0.3749 or 37.49%