Question

Irma bought 7 items at a back-to-school sale. The items cost:
$6.00 $8.00 $1.00 $2.00 $1.00 $3.00 $7.00
What was the standard deviation of the prices of the items?

If the answer is a decimal, round it to the nearest ten cents.

standard deviation (σ): $

Answer 1

To calculate the standard deviation of the prices of the items, we can follow these steps:

1. **Calculate the Mean (Average)**:
\[
\text{Mean} = \frac{\sum \text{prices}}{n}
\]
where \( n \) is the number of items.

The prices are: $6.00, $8.00, $1.00, $2.00, $1.00, $3.00, $7.00.

First, add the prices:
\[
6.00 + 8.00 + 1.00 + 2.00 + 1.00 + 3.00 + 7.00 = 28.00
\]
Then calculate the mean:
\[
\text{Mean} = \frac{28.00}{7} = 4.00
\]

2. **Calculate the Variance**:
The variance measures how much the prices deviate from the mean. The formula for variance (\( \sigma^2 \)) is:
\[
\sigma^2 = \frac{\sum (x_i - \mu)^2}{n}
\]
where \( x_i \) are the prices and \( \mu \) is the mean.

Now, we find the squared deviations from the mean:

- For $6.00: \quad (6.00 - 4.00)^2 = 2.00^2 = 4.00
- For $8.00: \quad (8.00 - 4.00)^2 = 4.00^2 = 16.00
- For $1.00: \quad (1.00 - 4.00)^2 = (-3.00)^2 = 9.00
- For $2.00: \quad (2.00 - 4.00)^2 = (-2.00)^2 = 4.00
- For $1.00: \quad (1.00 - 4.00)^2 = (-3.00)^2 = 9.00
- For $3.00: \quad (3.00 - 4.00)^2 = (-1.00)^2 = 1.00
- For $7.00: \quad (7.00 - 4.00)^2 = 3.00^2 = 9.00

Now, sum these squared deviations:
\[
4.00 + 16.00 + 9.00 + 4.00 + 9.00 + 1.00 + 9.00 = 52.00
\]

The variance is then:
\[
\sigma^2 = \frac{52.00}{7} \approx 7.42857
\]

3. **Calculate the Standard Deviation**:
The standard deviation (\( \sigma \)) is the square root of the variance:
\[
\sigma = \sqrt{7.42857} \approx 2.72699
\]

Rounding to the nearest ten cents:
\[
\sigma \approx 2.73
\]

Thus, the standard deviation of the prices of the items is approximately:

\[
\text{Standard deviation} (\sigma): \$2.73
\]