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 $100. Questions involve determining approximate percentages for buyers who paid specific price ranges (e.g., above $1300, below $800, etc.).

Solution

The problem relates to a normal distribution for the prices paid for an HD television, with the following given values:

  • Mean price (μ\mu) = $1000
  • Standard deviation (σ\sigma) = $100

The questions involve calculating approximate percentages for various ranges using the properties of the standard normal distribution (Z-scores).

Let’s break down each part:


1. What is the approximate percentage of buyers who paid more than $1300?

Z=13001000100=3Z = \frac{1300 - 1000}{100} = 3 From standard normal distribution tables, the probability corresponding to Z=3Z = 3 is approximately 0.99870.9987. The percentage beyond Z=3Z = 3 is: 10.9987=0.0013=0.13%1 - 0.9987 = 0.0013 = 0.13\% Answer: 0.13%


2. What is the approximate percentage of buyers who paid less than $800?

Z=8001000100=2Z = \frac{800 - 1000}{100} = -2 From standard normal distribution tables, the probability corresponding to Z=2Z = -2 is approximately 0.02280.0228. This gives: 2.28%2.28\% Answer: 2.28%


3. What is the approximate percentage of buyers who paid between $900 and $1100?

For 900900: Z=9001000100=1Z = \frac{900 - 1000}{100} = -1 The probability for Z=1Z = -1 is approximately 0.15870.1587.

For 11001100: Z=11001000100=1Z = \frac{1100 - 1000}{100} = 1 The probability for Z=1Z = 1 is approximately 0.84130.8413.

The percentage between $900 and $1100 is: 0.84130.1587=0.6826=68.26%0.8413 - 0.1587 = 0.6826 = 68.26\% Answer: 68.26%


4. What is the approximate percentage of buyers who paid between $900 and $1000?

For 900900 (Z=1Z = -1): P(Z<1)=0.1587P(Z < -1) = 0.1587 For 10001000 (Z=0Z = 0): P(Z<0)=0.5P(Z < 0) = 0.5

The percentage is: 0.50.1587=0.3413=34.13%0.5 - 0.1587 = 0.3413 = 34.13\% Answer: 34.13%


5. What is the approximate percentage of buyers who paid between $1000 and $1300?

For 10001000 (Z=0Z = 0): P(Z<0)=0.5P(Z < 0) = 0.5 For 13001300 (Z=3Z = 3): P(Z<3)=0.9987P(Z < 3) = 0.9987

The percentage is: 0.99870.5=0.4987=49.87%0.9987 - 0.5 = 0.4987 = 49.87\% Answer: 49.87%


6. What is the approximate percentage of buyers who paid between $1000 and $1200?

For 10001000 (Z=0Z = 0): P(Z<0)=0.5P(Z < 0) = 0.5 For 12001200 (Z=2Z = 2): P(Z<2)=0.9772P(Z < 2) = 0.9772

The percentage is: 0.97720.5=0.4772=47.72%0.9772 - 0.5 = 0.4772 = 47.72\% Answer: 47.72%


Let me know if you need further clarification or explanation!

Here are 5 relative questions to expand your understanding:

  1. What are the key properties of the standard normal distribution?
  2. How do you calculate a Z-score, and what does it represent?
  3. Why is the area under the curve in a normal distribution equal to 1?
  4. How would the results change if the standard deviation were larger (e.g., $200)?
  5. What is the relationship between empirical rules (68-95-99.7) and Z-scores?

Tip: When solving normal distribution problems, always sketch the curve and mark the regions you're calculating—this visual aid helps avoid errors!

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Math Problem Analysis

Mathematical Concepts

Statistics
Normal Distribution
Z-scores
Probability

Formulas

Z = (X - μ) / σ
P(Z < z) from standard normal tables

Theorems

Empirical Rule (68-95-99.7)
Properties of the Standard Normal Distribution

Suitable Grade Level

Grades 10-12