Question

A brewery's filling machine is adjusted to fill bottles with a mean of 32.7 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 32.73 oz? (Give your answer correct to four decimal places.)
(b) Let's say you buy 95 bottles of this ale for a party. How many bottles would you expect to find containing more than 32.73 oz. of ale? (Round your answer up to the nearest whole number.) bottles You may need to use the appropriate table in Appendix B to answer this question.

Answer 1

(a) P(X>32.73) = 0.2912

(b) Out of 95 bottles, 28 would contain more than 32.73 oz of ale.

Explanation:

(a) The amount of fill is normally distributed so we will calculate the z-score and then use it to find the probability using the normal distribution probability table.

Let the amount of fill be denoted by X. The z-score can be computed using the formula:

z = (X - μ)/σ

where μ = mean value of fill

σ = standard deviation of value of fill = √Variance

P(X>32.73) = 1 - P(X<32.73)

= 1 - P((X-μ)/σ < (32.73 - 32.7)/√0.003)

= 1 - P(z<0.55)

Using the normal distribution table in Appendix B, we can see the probability at z=0.55 is 0.7088. So,

P(X>32.73) = 1 - 0.7088

P(X>32.73) = 0.2912

(b) We are buying 95 bottles and we need to calculate how many of them contain more than 32.73 oz. For that, we will multiply the total number of bottles by the probability of finding more than 32.73 oz which we have calculated in (a).

95 * 0.2912 = 27.664

Rounding off to a whole number we get 28 bottles.

Out of 95 bottles, 28 would contain more than 32.73 oz of ale.