Solved on Nov 18, 2023

Using the normal distribution, find the percentage of buyers who paid between $\$14,000$ and $\$18,000$ for a car with mean $\$18,000$ and standard deviation $\$2,000$ .

STEP 1

Assumptions1. The distribution of car prices is normal. The mean price is $\$18,000$
3. The standard deviation is $\$,000$
4. We are using the68-95-99.7 Rule (also known as the empirical rule), which states that for a normal distribution, approximately68% of the data falls within one standard deviation of the mean,95% falls within two standard deviations, and99.7% falls within three standard deviations.

STEP 2

We need to find the number of standard deviations away from the mean the values $\$14,000$ and $\$18,000$ are. We can do this by subtracting the mean from each value and dividing by the standard deviation.
$Z = \frac{X - \mu}{\sigma}$ where $Z$ is the number of standard deviations, $X$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation.

STEP 3

Calculate the number of standard deviations for $\$14,000$ .
$Z_{14000} = \frac{\$14,000 - \$18,000}{\$2,000}$

STEP 4

Calculate the result.
$Z_{14000} = \frac{\$14,000 - \$18,000}{\$2,000} = -2$ This means that $\$14,000$ is2 standard deviations below the mean.

STEP 5

Calculate the number of standard deviations for $\$18,000$ .
$Z_{18000} = \frac{\$18,000 - \$18,000}{\$2,000}$

STEP 6

Calculate the result.
$Z_{18000} = \frac{\$18,000 - \$18,000}{\$2,000} =0$ This means that $\$18,000$ is0 standard deviations away from the mean, i.e., it is the mean.

STEP 7

Now, we need to use the68-95-99.7 Rule to find the percentage of buyers who paid between $\$14,000$ and $\$18,000$ . According to the rule,95% of the data falls within two standard deviations of the mean. Since $\$14,000$ is2 standard deviations below the mean and $\$18,000$ is the mean, we are looking at half of this95% range.

STEP 8

Calculate the percentage of buyers who paid between $\$14,000$ and $\$18,000$ .
$Percentage = \frac{95\%}{2}$

STEP 9

Calculate the result.
$Percentage = \frac{95\%}{2} =47.5\%$ The percentage of buyers who paid between $\$14,000$ and $\$18,000$ is47.5%.

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Using the normal distribution, find the percentage of buyers who paid between $\$14,000$ and $\$18,000$ for a car with mean $\$18,000$ and standard deviation $\$2,000$ .

STEP 1

STEP 2

STEP 3

Calculate the number of standard deviations for $\$14,000$ .
$Z_{14000} = \frac{\$14,000 - \$18,000}{\$2,000}$

STEP 4

Calculate the result.
$Z_{14000} = \frac{\$14,000 - \$18,000}{\$2,000} = -2$ This means that $\$14,000$ is2 standard deviations below the mean.

STEP 5

Calculate the number of standard deviations for $\$18,000$ .
$Z_{18000} = \frac{\$18,000 - \$18,000}{\$2,000}$

STEP 6

Calculate the result.
$Z_{18000} = \frac{\$18,000 - \$18,000}{\$2,000} =0$ This means that $\$18,000$ is0 standard deviations away from the mean, i.e., it is the mean.

STEP 7

STEP 8

Calculate the percentage of buyers who paid between $\$14,000$ and $\$18,000$ .
$Percentage = \frac{95\%}{2}$

STEP 9

Calculate the result.
$Percentage = \frac{95\%}{2} =47.5\%$ The percentage of buyers who paid between $\$14,000$ and $\$18,000$ is47.5%.

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Using the normal distribution, find the percentage of buyers who paid between $14,000\$14,000$14,000 and $18,000\$18,000$18,000 for a car with mean $18,000\$18,000$18,000 and standard deviation $2,000\$2,000$2,000.

STEP 1

STEP 2

STEP 3

Calculate the number of standard deviations for $14,000\$14,000$14,000.Z14000=$14,000−$18,000$2,000Z_{14000} = \frac{\$14,000 - \$18,000}{\$2,000}Z14000​=$2,000$14,000−$18,000​

STEP 4

Calculate the result.Z14000=$14,000−$18,000$2,000=−2Z_{14000} = \frac{\$14,000 - \$18,000}{\$2,000} = -2Z14000​=$2,000$14,000−$18,000​=−2This means that $14,000\$14,000$14,000 is2 standard deviations below the mean.

STEP 5

Calculate the number of standard deviations for $18,000\$18,000$18,000.Z18000=$18,000−$18,000$2,000Z_{18000} = \frac{\$18,000 - \$18,000}{\$2,000}Z18000​=$2,000$18,000−$18,000​

STEP 6

Calculate the result.Z18000=$18,000−$18,000$2,000=0Z_{18000} = \frac{\$18,000 - \$18,000}{\$2,000} =0Z18000​=$2,000$18,000−$18,000​=0This means that $18,000\$18,000$18,000 is0 standard deviations away from the mean, i.e., it is the mean.

STEP 7

STEP 8

Calculate the percentage of buyers who paid between $14,000\$14,000$14,000 and $18,000\$18,000$18,000.Percentage=95%2Percentage = \frac{95\%}{2}Percentage=295%​

STEP 9

Calculate the result.Percentage=95%2=47.5%Percentage = \frac{95\%}{2} =47.5\%Percentage=295%​=47.5%The percentage of buyers who paid between $14,000\$14,000$14,000 and $18,000\$18,000$18,000 is47.5%.

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Using the normal distribution, find the percentage of buyers who paid between $\$14,000$ and $\$18,000$ for a car with mean $\$18,000$ and standard deviation $\$2,000$ .

Calculate the number of standard deviations for $\$14,000$ .
$Z_{14000} = \frac{\$14,000 - \$18,000}{\$2,000}$

Calculate the result.
$Z_{14000} = \frac{\$14,000 - \$18,000}{\$2,000} = -2$ This means that $\$14,000$ is2 standard deviations below the mean.

Calculate the number of standard deviations for $\$18,000$ .
$Z_{18000} = \frac{\$18,000 - \$18,000}{\$2,000}$

Calculate the result.
$Z_{18000} = \frac{\$18,000 - \$18,000}{\$2,000} =0$ This means that $\$18,000$ is0 standard deviations away from the mean, i.e., it is the mean.

Calculate the percentage of buyers who paid between $\$14,000$ and $\$18,000$ .
$Percentage = \frac{95\%}{2}$

Calculate the result.
$Percentage = \frac{95\%}{2} =47.5\%$ The percentage of buyers who paid between $\$14,000$ and $\$18,000$ is47.5%.