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

1 answer

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 \( X \):

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

  1. Plug in the values:

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

\[ X = -128.16 + 949 \]

\[ X = 820.84 \]

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

Therefore, the answer is $821.