Question

A bottling plant fills one-gallon jugs with milk. The label on a one gallon jug of milk states that the volume of milk is 128 fluid ounces (fl.oz.) Federal law mandates that the jug must contain no less than the stated volume. The actual amount of milk in the jugs is normally distributed with mean µ = 129 fl. Oz. and standard deviation ơ = 0.8 fl. Oz. Use this information to answer below questions.

1. Find the z-score corresponding to a jug containing 128 fl. Oz. of milk?
2. What is the probability that a randomly selected jug will contain less than 128 fl. Oz. of milk?

Answer 1

1. -1.25

2. 0.1056

Explanation:

We are given that

mean=129

S.D=0.8

1.

The z-score can be found as

z=(x-mean)/S.D

z=(128-129)/0.8

z=-1/0.8

z=-1.25

Thus, a z-score corresponding to 128 fl. Oz. milk is -1.25.

2.

We have to find P(X<128).

A z-score is already found in part 1, So,

P(X<128)=P(Z<-1.25)

P(X<128)=P(-∞<Z<0)-P(0<Z<-1.25)

Using normal area table, we get P(0<Z<-1.25)=0.3944

P(X<128)=0.5-0.3944

P(X<128)=0.1056

Thus the probability that a randomly selected jug will contain less than 128 fl. Oz. of milk is 0.1056 or 10.56%