SOLUTION: The annual rainfall in a certain region is approximately normally distributed with mean 42.7 inches and standard deviation 5.7 inches. Round answers to the nearest tenth of a perce
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Question 1170929: The annual rainfall in a certain region is approximately normally distributed with mean 42.7 inches and standard deviation 5.7 inches. Round answers to the nearest tenth of a percent.
a) What percentage of years will have an annual rainfall of less than 44 inches? %
b) What percentage of years will have an annual rainfall of more than 40 inches? %
c) What percentage of years will have an annual rainfall of between 38 inches and 43 inches?
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's solve this problem using the properties of a normal distribution and a graphing calculator.
**Understanding the Problem**
* We have a normal distribution.
* Mean (μ) = 42.7 inches
* Standard deviation (σ) = 5.7 inches
**Using a Graphing Calculator**
As before, we'll use the `normalcdf` function:
* `normalcdf(lower_bound, upper_bound, mean, standard_deviation)`
**Calculations**
**a) Percentage of years with rainfall less than 44 inches**
* We want P(X < 44).
* `lower_bound`: -1E99 (or a very small number)
* `upper_bound`: 44
* `mean`: 42.7
* `standard_deviation`: 5.7
* `normalcdf(-1E99, 44, 42.7, 5.7)`
* Result: approximately 0.58987
* Percentage: 0.58987 \* 100% ≈ 59.0%
**b) Percentage of years with rainfall more than 40 inches**
* We want P(X > 40).
* `lower_bound`: 40
* `upper_bound`: 1E99 (or a very large number)
* `mean`: 42.7
* `standard_deviation`: 5.7
* `normalcdf(40, 1E99, 42.7, 5.7)`
* Result: approximately 0.68656
* Percentage: 0.68656 \* 100% ≈ 68.7%
**c) Percentage of years with rainfall between 38 inches and 43 inches**
* We want P(38 < X < 43).
* `lower_bound`: 38
* `upper_bound`: 43
* `mean`: 42.7
* `standard_deviation`: 5.7
* `normalcdf(38, 43, 42.7, 5.7)`
* Result: approximately 0.33924
* Percentage: 0.33924 \* 100% ≈ 33.9%
**Answers**
* a) 59.0%
* b) 68.7%
* c) 33.9%
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