Question

richards class jumped rope in p.e. he counted the number of jumps each student completed before getting tangled in the rope. the number of jumps 6 students completed was: 4 jumps 8 jumps 7 jumps 6 jumps 7 jumps 7 jumps what was the standard deviation of the numbers of jumps completed? if the answer is a decimal, round it to the nearest tenth. standard deviation (σ): jumps submit

Explanation:

Step1: Calculate the mean

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
Here \(n = 6\), \(x_1=4\), \(x_2 = 8\), \(x_3=7\), \(x_4=6\), \(x_5 = 7\), \(x_6=7\).
\(\bar{x}=\frac{4 + 8+7+6+7+7}{6}=\frac{39}{6}=6.5\)

Step2: Calculate \((x_{i}-\bar{x})^2\) for each \(x_{i}\)

• For \(x_1 = 4\): \((4 - 6.5)^2=(- 2.5)^2 = 6.25\)
• For \(x_2 = 8\): \((8 - 6.5)^2=(1.5)^2 = 2.25\)
• For \(x_3 = 7\): \((7 - 6.5)^2=(0.5)^2 = 0.25\)
• For \(x_4 = 6\): \((6 - 6.5)^2=(-0.5)^2 = 0.25\)
• For \(x_5 = 7\): \((7 - 6.5)^2=(0.5)^2 = 0.25\)
• For \(x_6 = 7\): \((7 - 6.5)^2=(0.5)^2 = 0.25\)

Step3: Calculate the variance \(\sigma^{2}\)

The formula for the variance \(\sigma^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^2}{n}\)
\(\sum_{i = 1}^{6}(x_{i}-\bar{x})^2=6.25+2.25 + 0.25+0.25+0.25+0.25=9.5\)
\(\sigma^{2}=\frac{9.5}{6}\approx1.583\)

Step4: Calculate the standard deviation \(\sigma\)

The formula for the standard deviation \(\sigma=\sqrt{\sigma^{2}}\)
\(\sigma=\sqrt{\frac{9.5}{6}}\approx1.3\)

Answer:

\(1.3\)