Math Problem Statement

The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is $1000 and the standard deviation is $120.

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What is the approximate percentage of buyers who paid less than $640? %

What is the approximate percentage of buyers who paid between $880 and $1120? %

What is the approximate percentage of buyers who paid between $1000 and $1360? %

What is the approximate percentage of buyers who paid between $880 and $1000? %

What is the approximate percentage of buyers who paid between $1000 and $1240? %

What is the approximate percentage of buyers who paid less than $760? %

Solution

This graph represents a normal distribution of the prices paid for a particular HD television model, with a mean (μ\mu) of $1000 and a standard deviation (σ\sigma) of $120. Given the nature of the normal distribution and the questions posed, we will apply properties of the empirical rule (also called the 68-95-99.7 rule) and the z-scores to find the percentages.

1. Percentage of buyers who paid less than $640

First, calculate the z-score: z=640−1000120=−360120=−3z = \frac{640 - 1000}{120} = \frac{-360}{120} = -3 A z-score of -3 corresponds to the far left of the distribution. Using standard normal distribution tables or the empirical rule, approximately 0.13% of buyers paid less than $640.

2. Percentage of buyers who paid between $880 and $1120

Find the z-scores:

  • For $880: z=880−1000120=−1z = \frac{880 - 1000}{120} = -1
  • For $1120: z=1120−1000120=1z = \frac{1120 - 1000}{120} = 1

According to the empirical rule, about 68% of the data falls within 1 standard deviation of the mean. Since this range includes the middle portion from z=−1z = -1 to z=1z = 1, 68% of buyers paid between $880 and $1120.

3. Percentage of buyers who paid between $1000 and $1360

Find the z-score for $1360: z=1360−1000120=3z = \frac{1360 - 1000}{120} = 3 From $1000 (mean) to $1360 (z = 3), about 49.87% of buyers fall in this range (since half of the buyers fall below the mean, and the area from z=0z = 0 to z=3z = 3 is about 49.87%).

4. Percentage of buyers who paid between $880 and $1000

The z-scores are the same as before:

  • $880: z=−1z = -1
  • $1000 (mean): z=0z = 0

From z=−1z = -1 to z=0z = 0, the area is half of the area for one standard deviation (from z=−1z = -1 to z=1z = 1). This corresponds to 34% of buyers paying between $880 and $1000.

5. Percentage of buyers who paid between $1000 and $1240

Find the z-score for $1240: z=1240−1000120=2z = \frac{1240 - 1000}{120} = 2 The area from z=0z = 0 to z=2z = 2 represents about 47.72% of buyers.

6. Percentage of buyers who paid less than $760

Find the z-score for $760: z=760−1000120=−2z = \frac{760 - 1000}{120} = -2 From z=−2z = -2, approximately 2.28% of buyers paid less than $760.

If you have any more questions or want more details, feel free to ask! Here are some related questions:

  1. What percentage of buyers paid more than $1360?
  2. What percentage of buyers paid between $760 and $1240?
  3. How do we interpret a z-score of zero in a normal distribution?
  4. Can you explain why the normal distribution is symmetrical?
  5. What percentage of buyers paid more than the mean price?

Tip: The area under the curve of a normal distribution always adds up to 100%, so when calculating percentages, subtracting from 100% can help find areas to the right of a z-score.

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Z-Scores
Empirical Rule (68-95-99.7 Rule)

Formulas

Z-Score: z = (X - μ) / σ
Percentage using Z-Scores

Theorems

Empirical Rule (68-95-99.7 Rule)

Suitable Grade Level

Grades 11-12

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