Descriptive Statistics Think of descriptive statistics as a way to tell the story of your data in a few key numbers. Instead of listing all 100 test scores, you might say: "the average is 75 and most scores are within 10 points of that." The mean shows where the center is, the median is the middle value (unaffected by extreme scores), and the mode is the most frequent. The standard deviation measures spread — small means data is tightly packed around the mean, large means it's widely scattered. Use n−1 (sample formula) for a subset of data, and N (population formula) when you have all the data.
$$\text{Mean: }\bar{x}=\frac{\sum x}{n}$$
$$\text{Sample Std Dev: }s=\sqrt{\frac{\sum (x-\bar{x})^2}{n-1}}$$
Find the mean, median, and mode of: 8, 10, 10, 12, 15, 17.
Show Solution $$\bar{x}=\frac{8+10+10+12+15+17}{6}=12$$
Median is the average of the 3rd and 4th values: $(10+12)/2=11$.
Mode is 10.
Sample Standard Deviation Find the sample standard deviation of 4, 6, 8, 10, 12.
Show Solution The mean is $8$.
$$\sum (x-\bar{x})^2=16+4+0+4+16=40$$
$$s=\sqrt{\frac{40}{4}}=3.162$$
Final answer: $3.16$.
Coefficient of Variation A data set has mean 80 and standard deviation 6. Find the coefficient of variation.
Show Solution $$CV=\frac{s}{\bar{x}}(100)=\frac{6}{80}(100)=7.5\%$$
Final answer: $7.5\%$.
Range and Population Variance Given the data set: 3, 7, 7, 19, find the range and population variance.
Show Solution Range = largest − smallest = 19 − 3 = 16 .
Mean: $\bar{x}=(3+7+7+19)/4=9$
$$\sigma^2=\frac{(3-9)^2+(7-9)^2+(7-9)^2+(19-9)^2}{4}=\frac{36+4+4+100}{4}=36$$
Final answer: Range = 16, Variance = 36.
Weighted Mean A student earned grades of 85, 90, and 78 in three subjects with credit units 3, 4, and 2 respectively. Find the weighted mean.
Show Solution Multiply each grade by its credit units, sum, then divide by total units.
$$\bar{x}_w=\frac{85(3)+90(4)+78(2)}{3+4+2}=\frac{255+360+156}{9}=\frac{771}{9}\approx85.67$$
Final answer: 85.67.
Grouped Data Mean Estimate the mean from this frequency table: Scores 60–69 (f = 5), 70–79 (f = 12), 80–89 (f = 8), 90–99 (f = 5). Total n = 30.
Show Solution Use class midpoints: 64.5, 74.5, 84.5, 94.5.
$$\bar{x}=\frac{64.5(5)+74.5(12)+84.5(8)+94.5(5)}{30}=\frac{322.5+894+676+472.5}{30}=\frac{2365}{30}\approx78.83$$
Final answer: 78.83.
Z-Score Calculation A dataset has mean 50 and standard deviation 10. Find the z-scores for values 65 and 35. Which is more extreme?
Show Solution $$z_{65}=\frac{65-50}{10}=+1.5, \quad z_{35}=\frac{35-50}{10}=-1.5$$
Both are equally extreme (1.5 standard deviations from the mean), just on opposite sides of the mean.
Variance Using the Shortcut Formula For the data 2, 4, 4, 4, 5, 5, 7, 9, find the population variance using $\sigma^2=\overline{x^2}-\bar{x}^2$.
Show Solution $n=8$, $\sum x=40$, so $\bar{x}=5$.
$\sum x^2=4+16+16+16+25+25+49+81=232$, so $\overline{x^2}=232/8=29$.
$$\sigma^2=29-5^2=29-25=4, \quad \sigma=2$$
Final answer: $\sigma^2=4$, $\sigma=2$.
Outlier Detection via Z-Score A dataset has mean 100 and standard deviation 15. A value of 148 is observed. Is it a potential outlier?
Show Solution $$z=\frac{148-100}{15}=3.2$$
A z-score with $|z|>3$ is generally considered a potential outlier. Since $z=3.2>3$, the value 148 is a potential outlier and should be investigated.
