Question

question 3
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 $mu = 50$ grams and standard deviation $sigma = 2$ 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 4 or more grams lighter than advertised?
a. it is extremely unlikely for this to occur; the probability is very close to 0.
b. there is about a 2.5% chance of this occurring.
c. there is about a 5% chance of this occurring.
d. there is no way to evaluate this likelihood, since the sample size ($n = 4$) is too small.
e. there is about a 16% chance of this occurring.

Explanation:

Step1: Identify the sampling - distribution of the sample mean

The sampling distribution of the sample mean $\bar{X}$ of a sample of size $n$ from a normal population with mean $\mu$ and standard deviation $\sigma$ is also normal, with mean $\mu_{\bar{X}}=\mu$ and standard deviation $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}$. Here, $\mu = 50$ grams, $\sigma = 2$ grams, and $n = 4$. So, $\sigma_{\bar{X}}=\frac{2}{\sqrt{4}}=\frac{2}{2}=1$ gram.

Step2: Calculate the z - score

We want to find the probability that the average weight of the bars in a pack is 4 or more grams lighter than advertised. The advertised weight is 50 grams, so we want to find $P(\bar{X}\leq50 - 4)=P(\bar{X}\leq46)$. The z - score is calculated as $z=\frac{\bar{X}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}$. Substituting $\bar{X} = 46$, $\mu_{\bar{X}}=50$, and $\sigma_{\bar{X}} = 1$ into the formula, we get $z=\frac{46 - 50}{1}=- 4$.

Step3: Find the probability using the standard normal distribution

The standard normal distribution table gives $P(Z\leq - 4)\approx0$.

Answer:

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