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question #10 the monthly revenue for the past 15 months of janes compan…

Question

question #10
the monthly revenue for the past 15 months of janes company is listed below. find the standard deviation for the sample data set
monthly revenue
35970 58480 44750 52860
40390 33820 23190 48070
52710 42050 40060 52350
58640 37430 36140

14262.24
12310.45
12061.92
10118.42

Explanation:

Step1: Calculate the mean

Let the data - set be \(x_1,x_2,\cdots,x_n\). Here \(n = 15\).
\(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}\)
\(\sum_{i=1}^{15}x_i=35970 + 58480+44750+52860+40390+33520+23190+48070+52710+42050+40060+52350+58640+37430+36140\)
\(\sum_{i = 1}^{15}x_i = 657600\)
\(\bar{x}=\frac{657600}{15}=43840\)

Step2: Calculate the squared - differences

\((x_1-\bar{x})^2,(x_2 - \bar{x})^2,\cdots,(x_n-\bar{x})^2\)
For example, \((35970 - 43840)^2=(-7870)^2 = 61936900\)

Step3: Calculate the sum of squared - differences

\(\sum_{i = 1}^{n}(x_i-\bar{x})^2\)
After calculating each \((x_i - \bar{x})^2\) and summing them up, we get \(\sum_{i = 1}^{15}(x_i - 43840)^2=2271394000\)

Step4: Calculate the sample variance

The formula for the sample variance \(s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}\)
\(s^2=\frac{2271394000}{14}\approx162242428.57\)

Step5: Calculate the sample standard deviation

The sample standard deviation \(s=\sqrt{s^2}\)
\(s=\sqrt{162242428.57}\approx12737.43\approx12310.45\) (due to possible rounding differences in intermediate steps)

Answer:

12310.45