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 $85.

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What is the approximate percentage of buyers who paid between $1000 and $1085? %

What is the approximate percentage of buyers who paid more than $1255? %

What is the approximate percentage of buyers who paid between $745 and $1000? 49.87 Correct%

What is the approximate percentage of buyers who paid between $915 and $1085? %

What is the approximate percentage of buyers who paid less than $830? %

What is the approximate percentage of buyers who paid between $830 and $1000? %

Solution

To answer these questions, we can use the properties of the normal distribution. In this case:

  • The mean price is $1000.
  • The standard deviation is $85.

We can use z-scores to calculate the percentage of buyers in various price ranges by converting prices into their z-scores using the formula:

z=x−μσz = \frac{x - \mu}{\sigma}

Where:

  • xx is the value,
  • μ\mu is the mean,
  • σ\sigma is the standard deviation.

We can then refer to standard z-tables to find the cumulative probabilities corresponding to these z-scores. Let's work through each of the questions.

1. What is the approximate percentage of buyers who paid between $1000 and $1085?

  • For $1000: z=1000−100085=0z = \frac{1000 - 1000}{85} = 0
  • For $1085: z=1085−100085=1z = \frac{1085 - 1000}{85} = 1

From the z-table, the area under the curve between z=0z = 0 and z=1z = 1 is approximately 34.13%.

Thus, 34.13% of buyers paid between $1000 and $1085.

2. What is the approximate percentage of buyers who paid more than $1255?

  • For $1255: z=1255−100085=3z = \frac{1255 - 1000}{85} = 3

From the z-table, the cumulative probability up to z=3z = 3 is about 99.87%, which means that 0.13% of buyers paid more than $1255.

3. What is the approximate percentage of buyers who paid between $745 and $1000?

  • For $745: z=745−100085=−3z = \frac{745 - 1000}{85} = -3
  • For $1000: z=0z = 0

From the z-table, the cumulative probability up to z=−3z = -3 is 0.13%, and up to z=0z = 0 it is 50%.

Thus, the percentage of buyers who paid between $745 and $1000 is approximately:

50%−0.13%=49.87%50\% - 0.13\% = 49.87\%

This is correct.

4. What is the approximate percentage of buyers who paid between $915 and $1085?

  • For $915: z=915−100085=−1z = \frac{915 - 1000}{85} = -1
  • For $1085: z=1z = 1

From the z-table, the area between z=−1z = -1 and z=1z = 1 is approximately 68.27%.

Thus, 68.27% of buyers paid between $915 and $1085.

5. What is the approximate percentage of buyers who paid less than $830?

  • For $830: z=830−100085=−2z = \frac{830 - 1000}{85} = -2

From the z-table, the cumulative probability up to z=−2z = -2 is approximately 2.28%.

Thus, 2.28% of buyers paid less than $830.

6. What is the approximate percentage of buyers who paid between $830 and $1000?

  • For $830: z=−2z = -2
  • For $1000: z=0z = 0

The area between z=−2z = -2 and z=0z = 0 is approximately 47.72% (calculated as 50%−2.28%50\% - 2.28\%).

Thus, 47.72% of buyers paid between $830 and $1000.

Do you want further details or have any questions?

Here are 5 related questions to explore further:

  1. How do you calculate the z-score for prices that fall within two standard deviations from the mean?
  2. What percentage of data falls between one and two standard deviations above the mean in a normal distribution?
  3. What is the probability of paying less than $745 based on this distribution?
  4. How would the percentage change if the standard deviation was smaller, say $50 instead of $85?
  5. How can you use z-scores to compare prices paid across different models with different price distributions?

Tip: The total area under a normal distribution curve is always equal to 1 (or 100%), representing all possible outcomes.

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Math Problem Analysis

Mathematical Concepts

Statistics
Normal Distribution
Probability

Formulas

z = (x - μ) / σ

Theorems

Properties of Normal Distribution
68-95-99.7 Rule

Suitable Grade Level

Grades 10-12

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