To solve this problem:

Given:

• The mean amount of sunscreen in a tube: $\mu = 496 \, \text{ml}$
• The standard deviation of the amount in a tube: $\sigma = 5 \, \text{ml}$
• Number of tubes in the sample: $n = 4$

Part (a): Standard deviation of the average, $\bar{x}$:

The formula for the standard deviation of the sample mean ($\bar{x}$) is:

$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

Substitute the values:

$\sigma_{\bar{x}} = \frac{5}{\sqrt{4}} = \frac{5}{2} = 2.5 \, \text{ml}$

So, the standard deviation of the average is 2.5 ml.

Part (b): Probability that the average amount of sunscreen from 4 tubes is less than 489 ml:

To calculate the probability, we first find the z-score for $\bar{x} = 489$.

The formula for the z-score is:

$z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}$

Substitute the values:

$z = \frac{489 - 496}{2.5} = \frac{-7}{2.5} = -2.8$

Using a standard normal distribution table or a calculator, the probability corresponding to $z = -2.8$ is approximately:

$P(Z < -2.8) \approx 0.0026$

So, the probability that the average amount of sunscreen is less than 489 ml is 0.26%.

Final Answers:

• (a) 2.5 ml
• (b) 0.0026 or 0.26%

Would you like further clarifications or details? Let me know if you need help understanding any steps!

Related Questions:

1. How does the sample size affect the standard deviation of the sample mean?
2. Why do we assume a normal distribution for this problem?
3. How would the probability change if the sample size increased to $n = 9$?
4. What is the probability that the average amount is greater than 500 ml?
5. How is the z-score related to standard deviation in this context?

Tip:

Always double-check if the sample size is large enough for normal approximation, especially when the population distribution is not explicitly normal!