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 = [tex] \frac{a+b}{2} [/tex]
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) = [tex] \frac{7.5+10.5}{2}= \frac{18}{2}= 9 [/tex]
Now, the standard deviation of the uniform distribution is given by the formula:
σ = [tex] \frac{b-a}{\sqrt{12}} [/tex]
So, the standard deviation(σ) = [tex] \frac{10.5-7.5}{\sqrt{12}}= \frac{3}{\sqrt{12}}= 0.8660254 [/tex]
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) = [tex] \frac{8.0 - 9}{0.8660254} =\frac{-1}{0.8660254}= -1.1547 [/tex]
and z(X= 9.0) = [tex] \frac{9.0 - 9}{0.8660254} = 0 [/tex]
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%
