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Part 1
In​ 2008, the per capita consumption of soft drinks in Country A was reported to be 19.24 gallons. Assume that the per capita consumption of soft drinks in Country A is approximately normally​ distributed, with a mean of 19.24 gallons and a standard deviation of 5 gallons. Complete parts​ (a) through​ (d) below.
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Part 1
a. What is the probability that someone in Country A consumed more than 12 gallons of soft drinks in​ 2008?
The probability is 
  
enter your response here.
​(Round to four decimal places as​ needed.)
Part 2
b. What is the probability that someone in Country A consumed between 4 and 6 gallons of soft drinks in​ 2008?
The probability is 
  
enter your response here.
​(Round to four decimal places as​ needed.)
Part 3
c. What is the probability that someone in Country A consumed less than 6 gallons of soft drinks in​ 2008?
The probability is 
  
enter your response here.
​(Round to four decimal places as​ needed.)
Part 4
d. 97​% of the people in Country A consumed less than how many gallons of soft​ drinks?
The probability is 97​% that someone in Country A consumed less than 
  
enter your response here gallons of soft drinks.
​(Round to two decimal places as​ needed.)

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.

In 2008, the per capita consumption of soft drinks in Country A was reported to be 19.24 gallons. Assume that the per capita consumption of soft drinks in Country A is approximately normally​ distributed, with a mean of 19.24 gallons and a standard deviation of 5 gallons. Complete parts (a) through (d):

(a) What is the probability that someone in Country A consumed more than 12 gallons of soft drinks in 2008?
(b) What is the probability that someone in Country A consumed between 4 and 6 gallons of soft drinks in 2008?
(c) What is the probability that someone in Country A consumed less than 6 gallons of soft drinks in 2008?
(d) 97% of the people in Country A consumed less than how many gallons of soft drinks?

This problem explores the normal distribution of soft drink consumption in Country A with a mean of 19.24 gallons and a standard deviation of 5 gallons. The question includes calculating probabilities for consumption levels, including more than 12 gallons and between 4 and 6 gallons, as well as determining the 97th percentile consumption level.

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