Calculating SD for 5 Test Scores

Find the standard deviation of these test scores: $72, 78, 80, 84, 86$

Find the mean: $\bar{x} = \frac{72 + 78 + 80 + 84 + 86}{5} = \frac{400}{5} = 80$ = Mean = $80$

Find each deviation from the mean: $72 - 80 = -8$ $78 - 80 = -2$ $80 - 80 = 0$ $84 - 80 = 4$ $86 - 80 = 6$ = Deviations: $-8, -2, 0, 4, 6$

Square each deviation: $(-8)^2 = 64$ $(-2)^2 = 4$ $(0)^2 = 0$ $(4)^2 = 16$ $(6)^2 = 36$ = Squared deviations: $64, 4, 0, 16, 36$

Find the variance (average of squared deviations): $\text{Variance} = \frac{64 + 4 + 0 + 16 + 36}{5} = \frac{120}{5} = 24$ = Variance = $24$

Take the square root: $\sigma = \sqrt{24} \approx 4.90$ = SD $\approx 4.90$

Answer: The standard deviation is approximately $4.90$ points. This means scores typically differ from the mean by about 5 points.

Comparing Spread of Two Data Sets

Data Set A: $10, 10, 10, 10, 10$ and Data Set B: $2, 6, 10, 14, 18$. Calculate SD for both.

Calculate mean for both sets: Set A: $\bar{x} = \frac{50}{5} = 10$ Set B: $\bar{x} = \frac{50}{5} = 10$ = Both means = $10$

Find deviations for Set A: All values are 10, so all deviations are: $10 - 10 = 0$ = Deviations A: $0, 0, 0, 0, 0$

Find deviations for Set B: $2-10=-8$, $6-10=-4$, $10-10=0$, $14-10=4$, $18-10=8$ = Deviations B: $-8, -4, 0, 4, 8$

Calculate variance for both: Set A: $\frac{0+0+0+0+0}{5} = 0$ Set B: $\frac{64+16+0+16+64}{5} = \frac{160}{5} = 32$ = Variance A = $0$, Variance B = $32$

Take square roots: $\sigma_A = \sqrt{0} = 0$ $\sigma_B = \sqrt{32} \approx 5.66$ = SD A = $0$, SD B $\approx 5.66$

Answer: Set A has SD = $0$ (no spread, all values identical). Set B has SD $\approx 5.66$ (values spread far from the mean). Same mean, very different spreads!

Sample vs Population SD

A sample of 4 temperatures: $18, 22, 20, 24$ degrees. Calculate the SAMPLE standard deviation.

Find the mean: $\bar{x} = \frac{18 + 22 + 20 + 24}{4} = \frac{84}{4} = 21$ = Mean = $21$

Find each deviation: $18 - 21 = -3$ $22 - 21 = 1$ $20 - 21 = -1$ $24 - 21 = 3$ = Deviations: $-3, 1, -1, 3$

Square each deviation: $(-3)^2 = 9$ $(1)^2 = 1$ $(-1)^2 = 1$ $(3)^2 = 9$ = Squared: $9, 1, 1, 9$

Find sample variance (divide by n-1): $s^2 = \frac{9 + 1 + 1 + 9}{4-1} = \frac{20}{3} \approx 6.67$ = Sample variance $\approx 6.67$

Take the square root: $s = \sqrt{6.67} \approx 2.58$ = Sample SD $\approx 2.58$

Answer: The sample standard deviation is approximately $2.58$ degrees. We used $n-1 = 3$ because this is a sample from a larger population of temperatures.

Mistake: Forgetting to square the deviations

Why: Without squaring, negative and positive deviations cancel out, giving a sum near zero regardless of actual spread.

Correct: Always square each deviation. The sum of squared deviations for a spread-out data set will be large.

Mistake: Dividing by $n-1$ instead of $n$ (or vice versa)

Why: Population SD uses $n$; sample SD uses $n-1$. Using the wrong formula gives incorrect results.

Correct: Use $n$ when you have all data (population). Use $n-1$ when you have a sample of a larger population.

Mistake: Forgetting to take the square root at the end

Why: Stopping at variance gives units that are squared (like 'points squared'), which is hard to interpret.

Correct: Always take the square root of variance to get standard deviation in the original units.

Mistake: Making arithmetic errors when summing squared deviations

Why: With many values, it's easy to make calculation mistakes.

Correct: Create a table with columns: Value, Deviation, Squared Deviation. Check each calculation.

Quality Control in Manufacturing

Factory managers calculate standard deviation to ensure products meet specifications consistently.

A factory produces bolts that should be $10$ mm long. If $\sigma = 0.1$ mm, 95% of bolts are within $\pm 0.2$ mm of target. If $\sigma = 0.5$ mm, there's too much variation.

Comparing Student Performance

Teachers calculate SD to understand how varied student scores are on tests.

If a class has a mean of 75 with SD = 5, most students score between 70 and 80. If SD = 15, scores range widely from 60 to 90.

Q: When do I use $n$ versus $n-1$ in the formula?

A: Use $n$ (population SD) when your data includes the entire group you're studying. Use $n-1$ (sample SD) when your data is a sample from a larger population. The $n-1$ correction gives a better estimate of the true population spread.

Q: Why do we square the deviations instead of using absolute values?

A: Squaring has mathematical advantages: it's differentiable (useful in calculus), gives more weight to larger deviations, and connects to the Pythagorean theorem and Euclidean distance. Absolute values work but don't have these properties.

Q: Can standard deviation ever be negative?

A: No! Since we're taking the square root of a sum of squared numbers, standard deviation is always zero or positive. SD = 0 only when all values are identical.