Question

question 5 of 11
a student weighs five different silver dollars. the resulting mass measurements are 26.37 g, 26.92 g, 26.30 g, 27.01 g, and 26.24 g. what is the average mass of the five silver dollars?
average mass: ​ g
what is the standard deviation of the measured masses?
standard deviation: ​ g

Explanation:

Part 1: Average Mass

Step1: Sum the masses

Sum the given masses: \(26.37 + 26.92 + 26.30 + 27.01 + 26.24\)
\[

$$\begin{align*} 26.37 + 26.92 &= 53.29\\ 53.29 + 26.30 &= 79.59\\ 79.59 + 27.01 &= 106.6\\ 106.6 + 26.24 &= 132.84 \end{align*}$$

\]

Step2: Divide by number of samples

There are 5 samples, so average \(=\frac{132.84}{5}\)
\[
\frac{132.84}{5} = 26.568
\]

Step1: Find the mean

We already found the mean (\(\bar{x}\)) is \(26.568\) g.

Step2: Calculate deviations from mean

For each mass \(x_i\), calculate \((x_i - \bar{x})^2\):

• For \(26.37\): \((26.37 - 26.568)^2 = (-0.198)^2 = 0.039204\)
• For \(26.92\): \((26.92 - 26.568)^2 = (0.352)^2 = 0.123904\)
• For \(26.30\): \((26.30 - 26.568)^2 = (-0.268)^2 = 0.071824\)
• For \(27.01\): \((27.01 - 26.568)^2 = (0.442)^2 = 0.195364\)
• For \(26.24\): \((26.24 - 26.568)^2 = (-0.328)^2 = 0.107584\)

Step3: Sum the squared deviations

Sum these values: \(0.039204 + 0.123904 + 0.071824 + 0.195364 + 0.107584\)
\[

$$\begin{align*} 0.039204 + 0.123904 &= 0.163108\\ 0.163108 + 0.071824 &= 0.234932\\ 0.234932 + 0.195364 &= 0.430296\\ 0.430296 + 0.107584 &= 0.53788 \end{align*}$$

\]

Step4: Divide by \(n - 1\) (for sample standard deviation)

\(n = 5\), so \(n - 1 = 4\). Then \(\frac{0.53788}{4} = 0.13447\)

Step5: Take the square root

Standard deviation \(= \sqrt{0.13447} \approx 0.3667\)

Answer:

\(26.57\) (rounded to two decimal places)

Part 2: Standard Deviation