Question

Suppose that a candy company makes a candy bar whose weight is supposed to be 50 grams, but in fact, the weight varies from bar to bar according to a normal distribution with mean μ = 50 grams and standard deviation σ = 3 grams.

If the company sells the candy bars in packs of 4 bars, what can we say about the likelihood that the average weight of the bars in a randomly selected pack is 6 or more grams lighter than advertised?


There is about a 16% chance of this occurring.

It is extremely unlikely for this to occur; the probability is very close to 0.

There is about a 2.5% chance of this occurring.

There is no way to evaluate this likelihood, since the sample size (n = 4) is too small.

There is about a 5% chance of this occurring.

Answer 1

Answer: To find the likelihood that the average weight of the bars in a randomly selected pack is 6 or more grams lighter than advertised, we need to consider the Central Limit Theorem.

According to the Central Limit Theorem, the sampling distribution of the sample mean (in this case, the average weight of the bars in a pack) will follow a normal distribution, regardless of the shape of the original population distribution, as long as the sample size is large enough (typically n ≥ 30).

Since the sample size is 4 bars, which is relatively small, we cannot directly apply the Central Limit Theorem. However, we can use the fact that the original population (individual candy bars) follows a normal distribution with mean μ = 50 grams and standard deviation σ = 3 grams.

For a sample size of 4, the standard deviation of the sample mean (also known as the standard error) is σ/√n = 3/√4 = 3/2 = 1.5 grams.

Now, we want to find the probability that the average weight of a randomly selected pack is 6 or more grams lighter than advertised. This is equivalent to finding the probability that the sample mean is less than or equal to 50 - 6 = 44 grams.

Next, we need to find the z-score for 44 grams:

z = (44 - 50) / 1.5 ≈ -4

Now, we look up the cumulative probability for a z-score of -4 in a standard normal distribution table (or use a calculator):

Cumulative probability ≈ 0.0000317

So, the likelihood that the average weight of the bars in a randomly selected pack is 6 or more grams lighter than advertised is about 0.00317%, which is extremely unlikely. The correct answer is:

It is extremely unlikely for this to occur; the probability is very close to 0.