Question

The cost to produce a batch of granola bars is approximately Normally distributed with a mean of $7.19 and a standard deviation of $0.86. If a random sample of 12 batches of granola bars is selected, what is the probability that the mean cost will be more than $7.00?

0.2220

0.2339

0.7653

0.7780

Answer 1

0.7780

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.

Population:

We have that
`\mu = 7.19, \sigma = 0.86`

Sample of 12:

`n = 12, s = (0.86)/(√(12)) = 0.2483`

What is the probability that the mean cost will be more than $7.00?

This is 1 subtracted by the pvalue of Z when X = 7. So

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

By the Central Limit Theorem

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

`Z = (7 - 7.19)/(0.2483)`

`Z = -0.765`

`Z = -0.765` has a pvalue of 0.222

1 - 0.222 = 0.778

So the answer is 0.7780