A machine fills 64-ounce jugs with detergent. Assume the distribution of the amount of detergent in these jugs is Normal. Under standard circumstances, the mean amount should be 64 ounces with a standard deviation of 0.4 ounces. A quality control inspector regularly checks the amount poured into the jugs to see if the machine needs an adjustment or not, which is needed when the machine either overflows or underfills the jugs. If the machine is running on target, what proportion of jugs receives more than 65 ounces of detergent

The proportion of jugs which receive more than 65 ounces of detergent is 0.0062 or 0.62%

Step-by-step explanation:

Mean amount of detergent = u = 64

Standard deviation = = 0.4

We need to find the proportion of jugs with over 65 ounces of detergent. Since the population is Normally Distributed and we have the value of population standard deviation, we will use the concept of z-score to solve this problem.

First we will convert 65 to its equivalent z-score, then using the z-table we will desired proportion. The formula to calculate the z-score is:

x = 65 converted to z score will be:

Therefore, the probability of detergent being more than 65 ounces is equivalent to probability of z-score being over 2.5

i.e.

P(X > 65) = P(z > 2.5)

From the z-table and using the property of symmetry:

P(z > 2.5) = 1 - P(z < 2.5)

= 1 - 0.9938

= 0.0062

Therefore,

P(X > 65) = P(z > 2.5) = 0.0062

So, the proportion of jugs which receive more than 65 ounces of detergent is 0.0062 or 0.62%