. In a certain normal distribution, 1.25% of the area lies to the left of 42, and 1.25% of the area lies to the right of 48. Find μ and σ

To find the mean (\(\mu\)) and standard deviation (\(\sigma\)) of the normal distribution, we can use the information given:

1. **1.25% of the area lies to the left of 42:** This implies that 42 is at the 1.25th percentile of the distribution.
2. **1.25% of the area lies to the right of 48:** This implies that 48 is at the 98.75th percentile of the distribution.

### Steps to Solve

1. **Determine the Z-scores:** Since 1.25% lies in the left tail and 1.25% in the right tail, we can find the Z-scores corresponding to these percentiles in a standard normal distribution.
   - For the 1.25th percentile (left tail), the Z-score is approximately \( Z = -2.24 \).
   - For the 98.75th percentile (right tail), the Z-score is approximately \( Z = +2.24 \).

2. **Set up equations using the Z-score formula:**
   The Z-score formula is:
   \[
   Z = \frac{X - \mu}{\sigma}
   \]
   where:
   - \( X \) is the value in the distribution,
   - \( \mu \) is the mean,
   - \( \sigma \) is the standard deviation.

   Since we have two values, 42 and 48, we set up two equations:

   - For \( X = 42 \) (1.25th percentile):
     \[
     -2.24 = \frac{42 - \mu}{\sigma}
     \]

   - For \( X = 48 \) (98.75th percentile):
     \[
     2.24 = \frac{48 - \mu}{\sigma}
     \]

3. **Solve for \(\mu\) and \(\sigma\):**
   We have a system of equations:
   \[
   -2.24 \sigma = 42 - \mu
   \]
   \[
   2.24 \sigma = 48 - \mu
   \]

   Adding these two equations to eliminate \(\mu\):
   \[
   -2.24 \sigma + 2.24 \sigma = 42 + 48 - 2\mu
   \]
   Simpl

In a certain normal distribution, 1.25% of the area lies to the left of 42, and 1.25% of the area lies to the right of 48. Find μ and σ.

Learn how to determine the mean (μ) and standard deviation (σ) of a normal distribution given that 1.25% of the area lies to the left of 42 and 1.25% lies to the right of 48. This solution covers normal distribution properties, percentiles, and Z-scores, suitable for Grades 11-12.

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