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Suppose that water usage in American showers is normally distributed, with an average shower using 18.9 gallons and a standard deviation of 3.1 gallons. Estimate the percentage of showers that used:

(a) between 12.7 and 25.1 gallons.
[tex]95.44\%[/tex]

(b) more than 25.1 gallons.
[tex]2.28\%[/tex]

(c) less than 9.6 gallons.
[tex]0.13\%[/tex]

(d) between 12.7 and 28.2 gallons.
[tex]97.59\%[/tex]



Answer :

GinnyAnswer
To solve the problem of water usages in American showers that are normally distributed, we need to determine the percentage of showers that fall within certain ranges based on the given mean and standard deviation. Let’s go step-by-step through each part of the question.

Given:
- Population mean ([tex]\(\mu\)[/tex]): 18.9 gallons
- Population standard deviation ([tex]\(\sigma\)[/tex]): 3.1 gallons

### (a) Between 12.7 and 25.1 gallons

1. Compute the z-scores for the lower and upper limits (12.7 and 25.1 gallons respectively):
[tex]\[ z_{\text{lower}} = \frac{12.7 - 18.9}{3.1} \][/tex]
[tex]\[ z_{\text{upper}} = \frac{25.1 - 18.9}{3.1} \][/tex]

2. Use the z-scores to find the cumulative probabilities from the standard normal distribution.

3. Subtract the cumulative probability of the lower z-score from that of the upper z-score to get the desired percentage.

The result is approximately:
[tex]\[ 95.44\% \][/tex]

### (b) More than 25.1 gallons

1. Compute the z-score for the lower limit (25.1 gallons):
[tex]\[ z_{\text{lower}} = \frac{25.1 - 18.9}{3.1} \][/tex]

2. Use the z-score to find the cumulative probability from the standard normal distribution.

3. Subtract this cumulative probability from 1 to get the desired percentage.

The result is approximately:
[tex]\[ 2.28\% \][/tex]

### (c) Less than 9.6 gallons

1. Compute the z-score for the upper limit (9.6 gallons):
[tex]\[ z_{\text{upper}} = \frac{9.6 - 18.9}{3.1} \][/tex]

2. Use the z-score to find the cumulative probability from the standard normal distribution.

The result is approximately:
[tex]\[ 0.13\% \][/tex]

### (d) Between 12.7 and 28.2 gallons

1. Compute the z-scores for the lower and upper limits (12.7 and 28.2 gallons respectively):
[tex]\[ z_{\text{lower}} = \frac{12.7 - 18.9}{3.1} \][/tex]
[tex]\[ z_{\text{upper}} = \frac{28.2 - 18.9}{3.1} \][/tex]

2. Use the z-scores to find the cumulative probabilities from the standard normal distribution.

3. Subtract the cumulative probability of the lower z-score from that of the upper z-score to get the desired percentage.

The result is approximately:
[tex]\[ 97.59\% \][/tex]

In summary:
- (a) The percentage of showers using between 12.7 and 25.1 gallons is approximately [tex]\(95.44\%\)[/tex].
- (b) The percentage of showers using more than 25.1 gallons is approximately [tex]\(2.28\%\)[/tex].
- (c) The percentage of showers using less than 9.6 gallons is approximately [tex]\(0.13\%\)[/tex].
- (d) The percentage of showers using between 12.7 and 28.2 gallons is approximately [tex]\(97.59\%\)[/tex].

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