Question

A 12 ounce bottle of beer contains 355 mL. The Random Brewing Company programs its machinery so that the amount of beer in a bottle is normally distributed with a mean of 355.5 mL and a standard deviation of 0.6 mL. It is reasonable to assume that amounts of beer filled into bottles are independent.Required:a. What is the probability that a randomly selected bottle has at least 355 ml?b. What is the probability that the bottles of a randomly selected 6-pack of beer have an average of at least 355 ml?c. What is the probability that the total amount of beer in the 6-pack is less than 2131.5 ml?

Answer 1

a) 79.67% probability that a randomly selected bottle has at least 355 ml

b) 99.29% probability that the bottles of a randomly selected 6-pack of beer have an average of at least 355 ml.

c) 88.88% probability that the total amount of beer in the 6-pack is less than 2131.5 ml

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question:

`\mu = 355.5, \sigma = 0.6`

a. What is the probability that a randomly selected bottle has at least 355 ml?

This is 1 subtracted by the pvalue of Z when X = 355. So

`Z = (X - \mu)/(\sigma)`

`Z = (355 - 355.5)/(0.6)`

`Z = -0.83`

`Z = -0.83` has a pvalue of 0.2033

1 - 0.2033 = 0.7967

79.67% probability that a randomly selected bottle has at least 355 ml.

b. What is the probability that the bottles of a randomly selected 6-pack of beer have an average of at least 355 ml?

Now
`n = 6, s = (0.5)/(√(6)) = 0.2041`

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (355 - 355.5)/(0.2041)`

`Z = -2.45`

`Z = -2.45` has a pvalue of 0.0071

1 - 0.0071 = 0.9929

99.29% probability that the bottles of a randomly selected 6-pack of beer have an average of at least 355 ml.

c. What is the probability that the total amount of beer in the 6-pack is less than 2131.5 ml?

2131.5/6 = 355.25

This is the pvalue of Z when X = 355.25.

`Z = (X - \mu)/(s)`

`Z = (355.25 - 355.5)/(0.2041)`

`Z = -1.22`

`Z = -1.22` has a pvalue of 0.1112

1 - 0.1112 = 0.8888

88.88% probability that the total amount of beer in the 6-pack is less than 2131.5 ml