Given the normal distribution of soft drink consumption with a mean of 19.24 gallons and a standard deviation of 5 gallons, we will solve each part step by step.

Part (a)

What is the probability that someone in Country A consumed more than 12 gallons of soft drinks in 2008?

1. First, calculate the Z-score for 12 gallons: $Z = \frac{X - \mu}{\sigma} = \frac{12 - 19.24}{5} = \frac{-7.24}{5} = -1.448$

2. Using the Z-score table, find the cumulative probability for $Z = -1.448$. This gives the probability that someone consumed less than 12 gallons. Looking up this Z-score in the standard normal table gives approximately 0.0739.

3. To find the probability of consuming more than 12 gallons: $P(X > 12) = 1 - P(X < 12) = 1 - 0.0739 = 0.9261$

So, the probability is 0.9261.

Part (b)

What is the probability that someone in Country A consumed between 4 and 6 gallons of soft drinks in 2008?

1. Find the Z-scores for 4 gallons and 6 gallons: $Z_4 = \frac{4 - 19.24}{5} = \frac{-15.24}{5} = -3.048$ $Z_6 = \frac{6 - 19.24}{5} = \frac{-13.24}{5} = -2.648$

2. Look up the cumulative probabilities for these Z-scores in the Z-table:

   • For $Z_4 = -3.048$, the cumulative probability is approximately 0.0011.
   • For $Z_6 = -2.648$, the cumulative probability is approximately 0.0041.

3. Subtract the cumulative probabilities: $P(4 < X < 6) = P(Z_6) - P(Z_4) = 0.0041 - 0.0011 = 0.0030$

So, the probability is 0.0030.

Part (c)

What is the probability that someone in Country A consumed less than 6 gallons of soft drinks in 2008?

From Part (b), we already have the cumulative probability for $Z = -2.648$, which corresponds to 6 gallons. So the probability is 0.0041.

Part (d)

97% of the people in Country A consumed less than how many gallons of soft drinks?

1. The cumulative probability is 0.97. We need to find the Z-score that corresponds to this cumulative probability. Using the Z-table, the Z-score for 0.97 is approximately 1.88.

2. Now, use the Z-score formula to find the corresponding value of X: $Z = \frac{X - \mu}{\sigma}$ $1.88 = \frac{X - 19.24}{5}$ $X - 19.24 = 1.88 \times 5 = 9.4$ $X = 19.24 + 9.4 = 28.64$

So, 97% of the people consumed less than 28.64 gallons of soft drinks.

---

Do you want further details or have any questions?

Here are 5 related questions for further practice:

1. What is the probability that someone consumed exactly 19.24 gallons of soft drinks?
2. If the standard deviation was reduced to 3 gallons, how would that change the probability in Part (a)?
3. How would you calculate the probability of someone consuming more than 25 gallons of soft drinks?
4. What is the Z-score for someone consuming 30 gallons of soft drinks?
5. What percentage of the population consumed between 15 and 20 gallons of soft drinks?

Tip: Always round your Z-scores and probabilities accurately based on the number of decimal places required.