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Difficulty: HardNormal Distributions, Standard Deviation Curves, and Percentile Ranks

The daily water consumption per household in a municipal district is normally distributed with a mean of 320320 liters and a standard deviation of 4040 liters. A local utility company classifies households into three consumption categories:

- Efficient: Daily consumption below 240240 liters
- Moderate: Daily consumption between 240240 liters and 400400 liters
- High: Daily consumption above 400400 liters

Based on the 68–95–99.768\text{--}95\text{--}99.7 empirical rule for normal distributions, in a random sample of 10,00010,000 households, how many more households are classified as Moderate than are classified as Efficient?

  1. A
    5,2005,200
  2. B
    7,9007,900
  3. C
    9,0009,000
  4. 9,2509,250Answer
  5. E
    9,5009,500

Answer

9,2509,250 households
The boundary values 240240 liters and 400400 liters represent z=−2z = -2 and z=+2z = +2, respectively. According to the empirical rule, 95%95\% of the population falls between these limits, yielding 9,5009,500 Moderate households. The remaining 5%5\% is split equally into the upper and lower tails (2.5%2.5\% each), meaning 2.5%2.5\% (250250 households) are Efficient. Subtracting 250250 from 9,5009,500 yields 9,2509,250.

Step-by-Step Solution

1
Calculate the z-scores for the boundaries of the Moderate and Efficient categories.
The boundary 240240 liters corresponds to z=240−32040=−2.0z = \frac{240 - 320}{40} = -2.0. The boundary 400400 liters corresponds to z=400−32040=+2.0z = \frac{400 - 320}{40} = +2.0.
Standardizing values into z-scores allows the direct application of the empirical rule.
2
Determine the percentage and count of households in the Moderate category.
By the 68–95–99.768\text{--}95\text{--}99.7 rule, 95%95\% of the data lies between z=−2.0z = -2.0 and z=+2.0z = +2.0. Count = 0.95×10,000=9,5000.95 \times 10,000 = 9,500 households.
The Moderate category spans from μ−2σ\mu - 2\sigma to μ+2σ\mu + 2\sigma.
3
Determine the percentage and count of households in the Efficient category.
The area below z=−2.0z = -2.0 is 100%−95%2=2.5%\frac{100\% - 95\%}{2} = 2.5\%. Count = 0.025×10,000=2500.025 \times 10,000 = 250 households.
The normal distribution is symmetric, so the remaining 5%5\% outer area is evenly split into two tails of 2.5%2.5\% each.
4
Calculate the difference between Moderate and Efficient household counts.
9,500−250=9,2509,500 - 250 = 9,250 households.
Subtracting the count of Efficient households from Moderate households satisfies the core question prompt.

Key Concept

Normal distribution z-score standardization and asymmetric region calculations using the 68-95-99.7 empirical rule.
Estimated Time:2m 30s
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Topics

Normal Distributions, Standard Deviation Curves, and Percentile Ranks

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