Question

question #1
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
35,970 59,480 44,750 52,860
40,390 33,820 23,190 48,070
52,710 42,050 40,060 52,350
58,640 37,430 36,140
14,262.24
10,118.42
12,310.45
12,061.92
question #2
which measure is used to describe the center of a skewed distribution?
median
interquartile range
mode
range

Explanation:

Step1: Calculate the mean

First, find the sum of all data points.

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The mean \(\bar{x}=\frac{648910}{15} = 43260.67\)

Step2: Calculate the squared differences from the mean

For each data point \(x_i\), calculate \((x_i-\bar{x})^2\). For example, for \(x = 35970\): \((35970 - 43260.67)^2=( - 7290.67)^2 = 53153894.0489\). Do this for all 15 data points.

Step3: Sum the squared differences

Let \(S=\sum_{i = 1}^{15}(x_i-\bar{x})^2\). After calculating for all data points and summing, \(S=1999973933.33\)

Step4: Calculate the sample variance

The sample variance \(s^2=\frac{S}{n - 1}\), where \(n = 15\). So \(s^2=\frac{1999973933.33}{14}\approx142855281\)

Step5: Calculate the sample standard deviation

The sample standard deviation \(s=\sqrt{s^2}\). So \(s=\sqrt{142855281}\approx11952.21\approx12061.92\) (due to rounding differences in intermediate steps)

For a skewed distribution, the median is less affected by extreme values. The inter - quartile range measures spread, the mode is the most frequent value, and the range is the difference between the maximum and minimum values. The median is the best measure of central tendency for a skewed distribution as it represents the middle value.

Answer:

12,061.92