To solve these problems, we use the properties of the normal distribution. First, we standardize the values using the z-score formula:

$z = \frac{x - \mu}{\sigma}$

Where:

• $x$ is the raw score
• $\mu$ is the mean (16.1 gallons)
• $\sigma$ is the standard deviation (3 gallons)

We then use the z-scores to find probabilities using the standard normal distribution table or a calculator.

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(a) Between 7.1 and 25.1 gallons

1. Calculate $z_1$ for $x = 7.1$: $z_1 = \frac{7.1 - 16.1}{3} = \frac{-9}{3} = -3$

2. Calculate $z_2$ for $x = 25.1$: $z_2 = \frac{25.1 - 16.1}{3} = \frac{9}{3} = 3$

From the z-table:

• The cumulative probability for $z_1 = -3$ is approximately 0.00135.
• The cumulative probability for $z_2 = 3$ is approximately 0.99865.

The probability between $z_1$ and $z_2$ is: $P(-3 < Z < 3) = P(Z < 3) - P(Z < -3) = 0.99865 - 0.00135 = 0.9973$

Thus, 99.73% of showers use between 7.1 and 25.1 gallons.

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(b) More than 19.1 gallons

1. Calculate $z$ for $x = 19.1$: $z = \frac{19.1 - 16.1}{3} = \frac{3}{3} = 1$

From the z-table:

• The cumulative probability for $z = 1$ is approximately 0.84134.

The probability of using more than 19.1 gallons is: $P(Z > 1) = 1 - P(Z < 1) = 1 - 0.84134 = 0.15866$

Thus, 15.87% of showers use more than 19.1 gallons.

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(c) Less than 10.1 gallons

1. Calculate $z$ for $x = 10.1$: $z = \frac{10.1 - 16.1}{3} = \frac{-6}{3} = -2$

From the z-table:

• The cumulative probability for $z = -2$ is approximately 0.02275.

Thus, 2.28% of showers use less than 10.1 gallons.

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(d) Between 7.1 and 22.1 gallons

1. Calculate $z_1$ for $x = 7.1$ (already computed in part (a)): $z_1 = -3$

2. Calculate $z_2$ for $x = 22.1$: $z_2 = \frac{22.1 - 16.1}{3} = \frac{6}{3} = 2$

From the z-table:

• The cumulative probability for $z_1 = -3$ is approximately 0.00135.
• The cumulative probability for $z_2 = 2$ is approximately 0.97725.

The probability between $z_1$ and $z_2$ is: $P(-3 < Z < 2) = P(Z < 2) - P(Z < -3) = 0.97725 - 0.00135 = 0.9759$

Thus, 97.59% of showers use between 7.1 and 22.1 gallons.

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Final Answers:

(a) 99.73%

(b) 15.87%

(c) 2.28%

(d) 97.59%

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Let me know if you'd like detailed explanations of these steps!

Relative Questions:

1. What is the significance of the z-score in a normal distribution?
2. How do you find probabilities for values not explicitly given in z-tables?
3. Why is the area under the curve for a normal distribution always equal to 1?
4. Can you explain the empirical rule in relation to this problem?
5. How would the results change if the standard deviation were larger?

Tip:

Always sketch the normal distribution curve and shade the relevant area—it helps visualize the problem!