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.

Question

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

Answer 1

a) Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.

b) Multiply the answer to a by 100.

Answer 2

They need 2 cup of flour.How many 1/4cups should they measure

Answer 3

To find the probability that the next randomly checked bottle contains more than 31.55 oz, we can use the normal distribution.

(a) Begin by finding the standard deviation (σ) using the variance given. The standard deviation is the square root of the variance:

σ = √(0.003) = 0.0549 (rounded to four decimal places)

Next, we calculate the z-score, which measures the number of standard deviations a particular value is from the mean. The formula for the z-score is:

z = (x - μ) / σ

where x is the value we're interested in (31.55 oz), μ is the mean (31.5 oz), and σ is the standard deviation (0.0549 oz).

Substituting the values, we get:

z = (31.55 - 31.5) / 0.0549 ≈ 0.9128

Using a normal distribution table or calculator, we find the probability corresponding to this z-score. The probability of finding a value greater than 31.55 oz is the area under the curve to the right of the z-score.

P(z > 0.9128) ≈ 0.1814 (rounded to four decimal places)

So, the probability that the next randomly checked bottle contains more than 31.55 oz is approximately 0.1814.

(b) To calculate the number of bottles you would expect to find containing more than 31.55 oz out of 100, we can use the probability found in part (a).

Expected number of bottles = probability * total number of bottles

Expected number of bottles = 0.1814 * 100 = 18.14

Rounded up to the nearest whole number, you would expect to find 19 bottles containing more than 31.55 oz of ale out of the 100 bottles.