Solution to Question 1:

Let’s break down and solve each part step-by-step:

Part (a)

The weight of the cereal boxes is normally distributed: $X \sim N(\mu, \sigma^2), \text{ where } \sigma = 5.$ The advertised weight is $500$ grams.

(i) Proportion of boxes underweight ($\mu = 505$):

The z-score is: $z = \frac{500 - \mu}{\sigma} = \frac{500 - 505}{5} = -1.$

Using the standard normal distribution table: $P(X < 500) = P(Z < -1) \approx 0.1587.$

Thus, 15.87% of boxes will be underweight when $\mu = 505$.

(ii) Find $\mu$ so only 1% of boxes are underweight ($P(X < 500) = 0.01$):

From the z-table, $P(Z < z) = 0.01$ corresponds to $z = -2.33$.

Using the z-score formula: $z = \frac{500 - \mu}{\sigma} \implies -2.33 = \frac{500 - \mu}{5}.$

Solving for $\mu$: $500 - \mu = -2.33 \cdot 5 \implies \mu = 500 + 11.65 = 511.65.$

Thus, $\mu = 511.65$ ensures that only 1% of boxes are underweight.

Part (b)

A sample of $n = 4$ boxes is taken. The sample mean weight, $\bar{X}$, is normally distributed: $\bar{X} \sim N(\mu, \frac{\sigma^2}{n}) = N(505, \frac{25}{4}).$

The standard deviation of $\bar{X}$ is: $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{5}{2} = 2.5.$

The z-score for $\bar{X} < 500$ is: $z = \frac{500 - \mu}{\sigma_{\bar{X}}} = \frac{500 - 505}{2.5} = -2.$

From the z-table: $P(\bar{X} < 500) = P(Z < -2) \approx 0.0228.$

Thus, the probability that the setting is changed is 2.28%.

Part (c)

A random sample of $n = 5$ boxes is taken. The probability of a single box being underweight ($P(X < 500)$) when $\mu = 505$ is $0.1587$ (from part (a)(i)).

The number of underweight boxes in the sample follows a Binomial distribution: $Y \sim B(n = 5, p = 0.1587).$

The machine setting is changed if more than one box is underweight, i.e., $P(Y > 1)$.

Using the complement rule: $P(Y > 1) = 1 - P(Y \leq 1).$

Compute $P(Y \leq 1)$: $P(Y \leq 1) = P(Y = 0) + P(Y = 1).$

For $Y = 0$: $P(Y = 0) = \binom{5}{0} p^0 (1-p)^5 = (0.8413)^5 \approx 0.418.$

For $Y = 1$: $P(Y = 1) = \binom{5}{1} p^1 (1-p)^4 = 5 \cdot (0.1587) \cdot (0.8413)^4 \approx 5 \cdot 0.1587 \cdot 0.499 \approx 0.396.$

Thus: $P(Y \leq 1) \approx 0.418 + 0.396 = 0.814.$

Finally: $P(Y > 1) = 1 - 0.814 = 0.186.$

The probability that the setting is changed is 18.6%.

Final Answers:

• (a)(i): $15.87\%$.
• (a)(ii): $\mu = 511.65$.
• (b): $2.28\%$.
• (c): $18.6\%$.

Do you need detailed steps or further clarification? Here are 5 questions for practice:

1. How would the z-score change if $\sigma = 10$?
2. How does increasing the sample size affect $\sigma_{\bar{X}}$?
3. Why does using a binomial distribution make sense in part (c)?
4. Can you generalize the formula for $\mu$ to ensure $k\%$ of boxes are underweight?
5. What assumptions are necessary for the normal and binomial approximations?

Tip: For small sample sizes ($n < 30$), always check if the population is approximately normal before using z-scores.