brewery's filling machine is adjusted to fill bottles with a mean of 33.8 oz. of ale and a variance of 0.004. Periodically, a bottle is checked and the amount of ale noted.

(a) Assuming the amount of fill is normally distributed, what is the probability that the next randomly checked bottle contains more than 33.88 oz?

I tried to plug in the info and square the variance. i have the Z Chart, but I am lost.

Z = (score-mean)/SD = (33.88-33.8)/SD

SD =√variance

To solve this problem, you can use the standard normal distribution and the Z-score formula. First, let's calculate the z-score for the given value of 33.88 oz using the formula:

Z = (X - μ) / σ

Where:
- X is the value we are interested in (33.88 oz)
- μ is the mean (33.8 oz)
- σ is the standard deviation (square root of variance = √0.004)

So, for the given problem:
X = 33.88 oz
μ = 33.8 oz
σ = √0.004 ≈ 0.0632

Now substitute these values into the formula and calculate the z-score:

Z = (33.88 - 33.8) / 0.0632

Z = 0.08 / 0.0632 ≈ 1.27

Now that we have the z-score, we can use the Z-chart (or Z-table) to find the corresponding probability. The z-score represents the number of standard deviations the value is away from the mean.

Looking up the z-score of 1.27 in the Z-chart, you will find that the corresponding probability is approximately 0.8980. However, since we are looking for the probability of the value being more than 33.88 oz, we need to find the probability of being to the right side of the z-score.

The Z-chart provides probabilities for the left side of the curve. To find the probability of being to the right side, subtract the value obtained from 1:

Probability = 1 - 0.8980 ≈ 0.1020 (or 10.20%)

Therefore, the probability that the next randomly checked bottle will contain more than 33.88 oz is approximately 0.1020 or 10.20%.