Question

find the standard deviation for the group of data items.

the standard deviation is (round to two decimal places as needed.)

Explanation:

Step1: Write out the data items

From the stem - and - leaf plot, the data items are \(x_1 = 37\), \(x_2=44\), \(x_3 = 47\), \(x_4=54\), \(x_5 = 57\). The number of data items \(n = 5\).

Step2: Calculate the mean \(\bar{x}\)

The formula for the mean is \(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}\).
\(\sum_{i=1}^{5}x_i=37 + 44+47+54+57=239\).
\(\bar{x}=\frac{239}{5}=47.8\).

Step3: Calculate \((x_i-\bar{x})^2\) for each \(x_i\)

• For \(x_1 = 37\): \((37 - 47.8)^2=(- 10.8)^2 = 116.64\)
• For \(x_2 = 44\): \((44 - 47.8)^2=(-3.8)^2=14.44\)
• For \(x_3 = 47\): \((47 - 47.8)^2=(-0.8)^2 = 0.64\)
• For \(x_4 = 54\): \((54 - 47.8)^2=(6.2)^2 = 38.44\)
• For \(x_5 = 57\): \((57 - 47.8)^2=(9.2)^2 = 84.64\)

Step4: Calculate the variance \(s^2\)

The formula for the variance (sample variance, but since the problem doesn't specify population or sample and for the sake of standard deviation calculation here, we can use the formula \(s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}\) (if it's a sample) or \(s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n}\) (if it's a population. Here we assume it's a population). \(\sum_{i = 1}^{n}(x_i-\bar{x})^2=116.64 + 14.44+0.64+38.44+84.64=254.8\).
\(s^2=\frac{254.8}{5}=50.96\).

Step5: Calculate the standard deviation \(s\)

The formula for the standard deviation is \(s=\sqrt{s^2}\).
\(s=\sqrt{50.96}\approx7.14\)

Answer:

\(7.14\)