Let’s solve the given problem step-by-step:

We are working with a normal distribution characterized by:

• Mean ($\mu$) = 160 g
• Standard deviation ($\sigma$) = 2 g

a) The percentage of tins containing over 164 g

We calculate the $Z$-score: $Z = \frac{X - \mu}{\sigma} = \frac{164 - 160}{2} = 2$ Using the standard normal distribution table, the area to the left of $Z = 2$ is 0.9772. The area to the right is: 1 - 0.9772 = 0.0228 \text{ (or 2.28%)}

b) The percentage of tins containing between 157 g and 163 g

For $X = 157$: $Z = \frac{157 - 160}{2} = -1.5$ From the table, the area to the left of $Z = -1.5$ is 0.0668.

For $X = 163$: $Z = \frac{163 - 160}{2} = 1.5$ From the table, the area to the left of $Z = 1.5$ is 0.9332.

The percentage of tins between 157 g and 163 g is: 0.9332 - 0.0668 = 0.8664 \text{ (or 86.64%)}

c) The probability that a tin contains less than 156 g

For $X = 156$: $Z = \frac{156 - 160}{2} = -2$ From the table, the area to the left of $Z = -2$ is: 0.0228 \text{ (or 2.28%)}

d) The minimum weight to ensure only 5% are underweight

We are looking for the value of $X$ such that the area to the left is 5% (or 0.05). From the table, this corresponds to: $Z = -1.645$ Rearranging the $Z$-score formula: $X = \mu + Z \cdot \sigma = 160 + (-1.645 \cdot 2) = 156.71 \, \text{g (approximately 156.7 g)}$

Final Answers:

a) $2.28\%$
b) $86.64\%$
c) $2.28\%$
d) $156.7 \, \text{g}$

Would you like me to elaborate further or clarify any part?
Here are 5 related questions you might find interesting:

1. How do we calculate probabilities for values not directly listed in the $Z$-score table?
2. What is the cumulative probability for a tin weighing exactly the mean weight?
3. How does increasing the standard deviation affect the distribution?
4. Can we use these techniques for non-normal distributions?
5. What does the $Z$-score represent intuitively?

Tip: Always verify the normality assumption of the data before applying these methods!