A brewery's filling machine is adjusted to fill bottles with a mean of 31.5 oz. of ale and a variance of 0.003. 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 31.55 oz? (Give your answer correct to four decimal places.)


(b) Let's say you buy 100 bottles of this ale for a party. How many bottles would you expect to find containing more than 31.55 oz. of ale? (Round your answer up to the nearest whole number.)
bottles

remember that variance = (sd)^2

sd = √.003 = .0548

z-score = (31.55-31.5)/.0548 = .91241
so from the table or graph we see
the BELOW 31.55 would be .8192
so above would be 1 - .8192 = .1808

b) number of bottles = .1808(100) = appr 18

thank you, but b part says it is wrong at 18 and I repeated what you done and got 18.08 which rounds to 18. so not sure why wrong?? but the formula will help with the next ones

There is a few of us working on these together so if you see different names that means none of us can figure it out. Again thank you

To solve this problem, we need to use the concept of the normal distribution and apply the properties of the mean and variance.

(a) To find the probability that the next randomly checked bottle contains more than 31.55 oz, we need to find the area under the normal curve to the right of 31.55 oz.

First, we calculate the standard deviation (σ) by taking the square root of the variance:
σ = √(0.003) = 0.0549

Next, we calculate the z-score, which is a measure of how many standard deviations away from the mean the value of 31.55 oz is:
z = (31.55 - 31.5) / 0.0549 ≈ 0.912

Using a z-table or a statistical software, we can look up the probability associated with a z-score of 0.912. The probability corresponds to the area under the normal curve to the right of 31.55 oz.

In this case, the probability is approximately 0.1806, when rounded to four decimal places.

(b) To find the number of bottles we would expect to find containing more than 31.55 oz out of 100 bottles, we can use the probability from part (a) and multiply it by the total number of bottles.

Expected number = (Probability) * (Total number of bottles)
Expected number = 0.1806 * 100

Rounding up to the nearest whole number, you would expect to find approximately 18 bottles containing more than 31.55 oz of ale out of the 100 bottles.