Question

The annual precipitation data of a city is normally distributed with mean \(1200\text{mm}\) and standard deviation \(200\text{mm}\), respectively. The probability that annual precipitation will be more than \(1400\text{mm}\)

• A. 0.16
• B. 0.68
• C. 0.84
• D. 0.75

Hint:

Remember the empirical rule (68-95-99.7 rule) for a normal distribution:
- About \(68%\) of the data lies within \(\pm 1\) standard deviation of the mean.
- This means the remaining \(32%\) lies in the tails outside this range (\(16%\) in each tail).
- Since \(1400\) is exactly 1 standard deviation above the mean, the probability of obtaining a value greater than 1400 is the area of the upper tail: \(16%\) or \(0.16\).
This allows you to find the answer quickly without using a standard normal table.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To find probabilities for any normal distribution, we first standardize the variable.
This converts our variable \(X\) into a standard normal variable \(Z\), which has a mean of 0 and a standard deviation of 1.

Step 2: Key Formula or Approach:
The formula to calculate the standard normal variable \(Z\) is:
\[ Z = \frac{X - \mu}{\sigma} \]
where \(\mu\) is the mean and \(\sigma\) is the standard deviation.

Step 3: Detailed Explanation:
Let \(X\) represent the annual precipitation of the city in millimeters.
We are given the parameters of the normal distribution:
- Mean, \(\mu = 1200\text{ mm}\)
- Standard deviation, \(\sigma = 200\text{ mm}\)
Therefore, we write: \(X \sim N(1200, 200^2)\).
We want to find the probability that the annual precipitation is greater than \(1400\text{ mm}\):
\[ P(X > 1400) \]
Now, standardize this value to find the corresponding \(Z\)-score:
\[ Z = \frac{1400 - 1200}{200} \]
\[ Z = \frac{200}{200} = 1 \]
This means a precipitation of \(1400\text{ mm}\) is exactly 1 standard deviation above the mean.
Our probability can now be written in terms of \(Z\):
\[ P(X > 1400) = P(Z > 1) \]
Using the complement rule:
\[ P(Z > 1) = 1 - P(Z \le 1) = 1 - \Phi(1) \]
where \(\Phi(1)\) is the cumulative distribution function of the standard normal distribution at \(Z = 1\).
From standard normal distribution tables, the cumulative probability at \(Z = 1\) is approximately:
\[ \Phi(1) \approx 0.8413 \]
Substitute this value back into our equation:
\[ P(Z > 1) = 1 - 0.8413 = 0.1587 \]
Rounding to two decimal places, we get:
\[ P(X > 1400) \approx 0.16 \]
This matches the first option.

Step 4: Final Answer:
Therefore, the correct option is (A).