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 μ = 50 grams and standard deviation σ = 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. there is no way to evaluate this likelihood, since the sample size (n = 4) is too small.
b. there is about a 2.5% chance of this occurring.
c. there is about a 16% chance of this occurring.
d. it is extremely unlikely for this to occur; the probability is very close to 0.
e. there is about a 5% chance of this occurring.

Explanation:

Step1: Identify the sampling - distribution parameters

The population mean $\mu = 50$ grams and the population standard deviation $\sigma=2$ grams. The sample size $n = 4$. The mean of the sampling - distribution of the sample mean is $\mu_{\bar{x}}=\mu = 50$ grams, and the standard deviation of the sampling - distribution of the sample mean (also known as the standard error) is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{4}} = 1$ gram.

Step2: Calculate the z - score

We want to find the probability that the sample - mean $\bar{x}$ is $4$ or more grams lighter than the advertised weight of $50$ grams, i.e., $\bar{x}\leq46$. The z - score is calculated using the formula $z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}$. 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

We want to find $P(\bar{X}\leq46)=P(Z\leq - 4)$. Looking up the value in the standard - normal table (or using a calculator with a normal - distribution function), the area to the left of $z=-4$ under the standard - normal curve is extremely small. The probability that $Z\leq - 4$ is very close to $0$.

Answer:

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