QuestionFind the percentage of buyers who paid between \ $22,500 and \$24,500, given a mean of \$22,500 and std dev of \$1000. The percentage is$ \square \%$.

Studdy Solution

STEP 1

Assumptions1. The mean price of the car is $22,500. . The standard deviation of the prices is$ 1,000.
3. The prices follow a normal distribution.
4. We are using the68-95-99.7 Rule, which states that in 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

First, we need to determine how many standard deviations away from the mean the price of $24,500 is. We can do this by subtracting the mean from the given price and then dividing by the standard deviation.
$Z = \frac{X - \mu}{\sigma}$

STEP 3

Now, plug in the given values for X (the given price), μ (the mean), and σ (the standard deviation) to calculate the Z-score.
$Z = \frac{\$24,500 - \$22,500}{\$1,000}$

STEP 4

Calculate the Z-score.
$Z = \frac{\$24,500 - \$22,500}{\$1,000} =2$

STEP 5

The Z-score of2 means that the price of $24,500 is two standard deviations away from the mean. According to the68-95-99.7 Rule, approximately95% of the data falls within two standard deviations of the mean. However, this includes both sides of the mean.

STEP 6

Since we are only interested in the percentage of buyers who paid between the mean and two standard deviations above the mean, we need to consider only one side of the distribution. Therefore, we divide the percentage by2.
$Percentage = \frac{95\%}{2}$

STEP 7

Calculate the percentage.
$Percentage = \frac{95\%}{2} =47.5\%$ Approximately47.5% of buyers paid between $22,500 and$ 24,500 for the car.

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QuestionFind the percentage of buyers who paid between \ $22,500 and \$24,500, given a mean of \$22,500 and std dev of \$1000. The percentage is$ \square \%$.

Studdy Solution

STEP 1

STEP 2

First, we need to determine how many standard deviations away from the mean the price of $24,500 is. We can do this by subtracting the mean from the given price and then dividing by the standard deviation.
$Z = \frac{X - \mu}{\sigma}$

STEP 3

Now, plug in the given values for X (the given price), μ (the mean), and σ (the standard deviation) to calculate the Z-score.
$Z = \frac{\$24,500 - \$22,500}{\$1,000}$

STEP 4

Calculate the Z-score.
$Z = \frac{\$24,500 - \$22,500}{\$1,000} =2$

STEP 5

The Z-score of2 means that the price of $24,500 is two standard deviations away from the mean. According to the68-95-99.7 Rule, approximately95% of the data falls within two standard deviations of the mean. However, this includes both sides of the mean.

STEP 6

Since we are only interested in the percentage of buyers who paid between the mean and two standard deviations above the mean, we need to consider only one side of the distribution. Therefore, we divide the percentage by2.
$Percentage = \frac{95\%}{2}$

STEP 7

Calculate the percentage.
$Percentage = \frac{95\%}{2} =47.5\%$ Approximately47.5% of buyers paid between $22,500 and$ 24,500 for the car.

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QuestionFind the percentage of buyers who paid between \22,500and$24,500,givenameanof$22,500andstddevof$1000.Thepercentageis22,500 and \$24,500, given a mean of \$22,500 and std dev of \$1000. The percentage is 22,500and$24,500,givenameanof$22,500andstddevof$1000.Thepercentageis\square \%$.

Studdy Solution

STEP 1

STEP 2

First, we need to determine how many standard deviations away from the mean the price of $24,500 is. We can do this by subtracting the mean from the given price and then dividing by the standard deviation.Z=X−μσZ = \frac{X - \mu}{\sigma}Z=σX−μ​

STEP 3

Now, plug in the given values for X (the given price), μ (the mean), and σ (the standard deviation) to calculate the Z-score.Z=$24,500−$22,500$1,000Z = \frac{\$24,500 - \$22,500}{\$1,000}Z=$1,000$24,500−$22,500​

STEP 4

Calculate the Z-score.Z=$24,500−$22,500$1,000=2Z = \frac{\$24,500 - \$22,500}{\$1,000} =2Z=$1,000$24,500−$22,500​=2

STEP 5

The Z-score of2 means that the price of $24,500 is two standard deviations away from the mean. According to the68-95-99.7 Rule, approximately95% of the data falls within two standard deviations of the mean. However, this includes both sides of the mean.

STEP 6

Since we are only interested in the percentage of buyers who paid between the mean and two standard deviations above the mean, we need to consider only one side of the distribution. Therefore, we divide the percentage by2.Percentage=95%2Percentage = \frac{95\%}{2}Percentage=295%​

STEP 7

Calculate the percentage.Percentage=95%2=47.5%Percentage = \frac{95\%}{2} =47.5\%Percentage=295%​=47.5%Approximately47.5% of buyers paid between 22,500and22,500 and 22,500and24,500 for the car.

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QuestionFind the percentage of buyers who paid between \ $22,500 and \$24,500, given a mean of \$22,500 and std dev of \$1000. The percentage is$ \square \%$.

First, we need to determine how many standard deviations away from the mean the price of $24,500 is. We can do this by subtracting the mean from the given price and then dividing by the standard deviation.
$Z = \frac{X - \mu}{\sigma}$

Now, plug in the given values for X (the given price), μ (the mean), and σ (the standard deviation) to calculate the Z-score.
$Z = \frac{\$24,500 - \$22,500}{\$1,000}$

Calculate the Z-score.
$Z = \frac{\$24,500 - \$22,500}{\$1,000} =2$

Since we are only interested in the percentage of buyers who paid between the mean and two standard deviations above the mean, we need to consider only one side of the distribution. Therefore, we divide the percentage by2.
$Percentage = \frac{95\%}{2}$

Calculate the percentage.
$Percentage = \frac{95\%}{2} =47.5\%$ Approximately47.5% of buyers paid between $22,500 and$ 24,500 for the car.