In City A, monthly rainfall was recorded to be normally distributed with mean 226 mm and standard deviation 52 mm .

Given that the monthly rainfall in City A follows a normal distribution with a **mean (μ)** of 226 mm and a **standard deviation (σ)** of 52 mm, you can calculate various probabilities and percentiles for this distribution.

For example, with a normal distribution:

1. **68% of data** lies within 1 standard deviation of the mean, i.e., between:
   \[
   \mu - \sigma = 226 - 52 = 174 \, \text{mm}
   \quad \text{and} \quad
   \mu + \sigma = 226 + 52 = 278 \, \text{mm}
   \]
   
2. **95% of data** lies within 2 standard deviations, i.e., between:
   \[
   \mu - 2\sigma = 226 - 2(52) = 122 \, \text{mm}
   \quad \text{and} \quad
   \mu + 2\sigma = 226 + 2(52) = 330 \, \text{mm}
   \]

3. **99.7% of data** lies within 3 standard deviations, i.e., between:
   \[
   \mu - 3\sigma = 226 - 3(52) = 70 \, \text{mm}
   \quad \text{and} \quad
   \mu + 3\sigma = 226 + 3(52) = 382 \, \text{mm}
   \]

Do you want to calculate any specific probabilities, like how likely it is for the rainfall to exceed or fall below a certain value? Let me know if you'd like more details or clarifications on how to proceed with further calculations!

Here are some related questions you might explore:
1. What is the probability that the rainfall is less than 200 mm in a month?
2. What is the probability that rainfall is

In City A, monthly rainfall was recorded to be normally distributed with mean 226 mm and standard deviation 52 mm.

Explore the statistical analysis of monthly rainfall in City A, where the data follows a normal distribution with a mean of 226 mm and a standard deviation of 52 mm. Learn how to apply the 68-95-99.7 rule to understand the distribution of rainfall, suitable for students in Grades 10-12.

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