The daily water consumption for an Ohio community is normally distributed with a mean consumption of 800,000 gallons and a standard deviation of 80,000 gallons. The community water system will experience a noticeable drop in water pressure when the daily water consumption exceeds 984,000 gallons. What is the probability of experiencing such a drop in water pressure?
Solution
Let $X$ denote the daily water consumption for an Ohio community.
The daily water consumption for an Ohio community is normally distributed with a mean consumption of 800,000 gallons and a standard deviation of 80,000 gallons.
$X\sim N(800000,80000^2)$. That $\mu=800,000$ gallons and $\sigma = 80,000$ gallons.
The community water system will experience a noticeable drop in water pressure when the daily water consumption exceeds 984,000 gallons.
Thus the probability of experiencing such a drop in water pressure is
$$ \begin{aligned} P(X \geq 984000) &= 1-P(X < 984000)\\ &= 1- P\bigg(\frac{X-\mu}{\sigma} < \frac{984000-800000}{80000}\bigg)\\ &=1- P\bigg(Z < 2.3\bigg)\\ &=1-0.9893\\ &= 0.0107 \end{aligned} $$
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
