Question

In a distribution exactly normal, 7% of the items are under 35 and 89% are under 63. Find the mean and standard deviation of the distribution.

Hint:

In normal distribution problems, use z-scores to relate the given percentages to the mean and standard deviation. Use the z-score table for accurate values.

Answer 1

Step 1: Understanding the normal distribution.
We are given a normal distribution with certain percentages under specific values. We need to use the z-scores to relate these percentages to the mean and standard deviation.
Step 2: Use of z-scores.
For a normal distribution: - The z-score formula is: \[ z = \frac{x - \mu}{\sigma} \] where \( x \) is the data value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
Step 3: Find the z-scores corresponding to the given percentages.
- For 7%, we use the z-score table to find that the z-score corresponding to 7% in the left tail is approximately \( z = -1.475 \). - For 89%, the z-score corresponding to 89% is approximately \( z = 1.23 \).
Step 4: Set up two equations.
From the z-score formula for both 35 and 63: \[ \frac{35 - \mu}{\sigma} = -1.475 \quad \text{(1)} \] \[ \frac{63 - \mu}{\sigma} = 1.23 \quad \text{(2)} \]
Step 5: Solve the system of equations.
- From equation (1), we get: \[ 35 - \mu = -1.475\sigma \quad \Rightarrow \quad \mu = 35 + 1.475\sigma \] - Substitute this value of \( \mu \) into equation (2): \[ \frac{63 - (35 + 1.475\sigma)}{\sigma} = 1.23 \] Simplifying this equation, we can solve for \( \sigma \), and then substitute back to find \( \mu \).
Step 6: Conclusion.
After solving, we find: \[ \mu = 50 \quad \text{and} \quad \sigma = 10 \]