Question

The average price of a personal computer is $949. If the computer prices are normally distributed with a standard deviation of $100, the least expensive 10% of personal computers cost less than what amount? Round your answer to the nearest dolla

Answer 1

To find the value below which the least expensive 10% of personal computers fall, we need to determine the corresponding z-score that represents the 10th percentile in a standard normal distribution.

1. **Identify the z-score for the 10th percentile:**
From z-tables or using a standard normal distribution calculator, the z-score that corresponds to the 10th percentile is approximately \(-1.2816\).

2. **Use the z-score formula to find the value (X):**
The z-score formula is:

\[
z = \frac{X - \mu}{\sigma}
\]

Where:
- \( z \) is the z-score.
- \( \mu \) is the mean (average price).
- \( \sigma \) is the standard deviation.
- \( X \) is the value we want to find.

Given:
- \( \mu = 949 \)
- \( \sigma = 100 \)
- \( z = -1.2816 \)

We can rearrange the formula to solve for \( X \):

\[
X = z \cdot \sigma + \mu
\]

3. **Plug in the values:**

\[
X = -1.2816 \cdot 100 + 949
\]

\[
X = -128.16 + 949
\]

\[
X = 820.84
\]

4. **Round to the nearest dollar:**
The least expensive 10% of personal computers cost less than approximately $821.

Therefore, the answer is **$821**.