Standard deviation and variance worksheet with worked answers
This worksheet contains ten exercises, from finding a mean to interpreting a standard deviation. The first few take a couple of minutes by hand; later ones bring in the shortcut formula, the difference between a population and a sample, frequency tables and grouped classes. Every exercise has a full worked answer.
It is written for high-school students revising statistics and for teachers who want a sheet they can print or adapt.
A4 PDF: questions first, answers on separate pages.
Licence and reuse
This worksheet is released under the Creative Commons Attribution 4.0 International licence (CC BY 4.0). You may copy, share, adapt and reuse it, in class, on a learning platform or in a textbook, as long as you credit the source.
Suggested credit: “Standard deviation and variance worksheet, Rédaction MaCalculatrice, macalculatriceenligne.com, CC BY 4.0”. If you change it, say so.
Key formulas
- Mean of N values: x = (x1 + x2 + … + xN) ÷ N.
- Variance (population): V = [(x1 − x)² + … + (xN − x)²] ÷ N.
- Standard deviation: σ = √V, in the same unit as the data.
- Shortcut form: V = (x1² + … + xN²) ÷ N − x².
- Sample: to estimate the spread of a population from a sample, divide by N − 1 instead of N: s² = [(x1 − x)² + … + (xN − x)²] ÷ (N − 1).
- Data with frequencies ni: x = Σ nixi ÷ N and V = Σ nixi² ÷ N − x², where N = Σ ni.
Rounding: a result written with “=” is exact; a result written with “≈” is rounded to 2 decimal places (a 5 in the third decimal place rounds up). Intermediate steps use exact values.
Questions
Exercise 1. Finding a mean
Eight students sat a test marked out of 20. Their marks were: 14, 7, 8, 13, 18, 9, 11, 16.
Find the mean mark x.
Exercise 2. Variance of a small data set
Over one term, five students borrowed this many books from the school library: 12, 14, 6, 8, 15.
Only these five students are studied (they are the whole population). Find the mean, then the variance V.
Exercise 3. Standard deviation and units
Daily maximum temperatures recorded in a town over six days (°C): 8, 17, 9, 13, 11, 14.
a) Find the variance V and the standard deviation σ (treat the six days as the population).
b) What unit is V measured in? What about σ?
Exercise 4. The shortcut formula (mean of squares)
Daily step counts (in thousands) for one person over five days: 6, 10, 3, 12, 10.
Find the variance using V = (mean of the squares) − x², then the standard deviation.
Exercise 5. Population or sample: divide by N or by N − 1
A basketball club has several hundred members. Six players are picked at random and their heights are measured (cm): 172, 195, 184, 171, 190, 180.
a) Find the standard deviation σ of these six heights, dividing by N.
b) Find the standard deviation s, dividing by N − 1.
c) Which of the two should be used to estimate the spread of heights across the whole club?
Exercise 6. Grouped data: frequency table
Students in one class were asked how many books they read last month:
| Books read | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Number of students | 8 | 1 | 4 | 1 | 7 | 4 |
Find the total number of students N, then the mean, variance and standard deviation of the number of books read.
Exercise 7. Data grouped into classes
Travel time from home to school for 40 students, grouped into classes:
| Time (min) | Students |
|---|---|
| [0, 10) | 12 |
| [10, 20) | 7 |
| [20, 30) | 5 |
| [30, 40) | 8 |
| [40, 60) | 8 |
a) Why is it impossible to find the exact mean?
b) Using the midpoint of each class, estimate the mean and the standard deviation.
Exercise 8. Interpreting: same mean, different spread
Marks out of 20 for eight students in each of two groups:
Group A: 10, 11, 12, 12, 12, 12, 13, 14.
Group B: 7, 7, 9, 10, 14, 15, 17, 17.
a) Show that both groups have the same mean.
b) Find the standard deviation of each group.
c) Which group is more consistent? Write a one-sentence conclusion.
Exercise 9. What happens when every value changes
Price of the same notebook in five shops (€): 14, 13, 6, 10, 12.
a) Find the mean and the standard deviation of these prices.
b) Every shop raises its price by €2. Give the new mean and the new standard deviation without redoing the whole calculation.
c) Starting again from the original prices, every shop raises its price by 10%. Same question.
Exercise 10. The “mean ± 2 standard deviations” rule
Mass (g) of the 20 apples in one crate: 171, 150, 172, 215, 162, 140, 168, 157, 160, 145, 154, 171, 156, 169, 167, 162, 167, 145, 152, 150.
You are given Σxi = 3233 and Σxi² = 527397.
a) Find the mean and the standard deviation (the whole crate is the population).
b) Work out the bounds of the interval [x − 2σ, x + 2σ].
c) How many apples have a mass inside this interval? What percentage is that? What can you say about the remaining apple?
Worked answers
Exercise 1: Show the worked answer
1. Add the marks: 14 + 7 + 8 + 13 + 18 + 9 + 11 + 16 = 96.
2. Count the values: N = 8.
3. x = 96 ÷ 8 = 12.
Answer: the mean mark is 12 out of 20.
Exercise 2: Show the worked answer
1. Mean: x = (12 + 14 + 6 + 8 + 15) ÷ 5 = 55 ÷ 5 = 11.
2. Work out how far each value is from the mean, then square it:
| xi | xi − x | (xi − x)² |
|---|---|---|
| 12 | 1 | 1 |
| 14 | 3 | 9 |
| 6 | −5 | 25 |
| 8 | −3 | 9 |
| 15 | 4 | 16 |
| Total | 0 | 60 |
Check: the deviations always add up to 0, here 0.
3. V = sum of squares ÷ N = 60 ÷ 5 = 12.
Answer: x = 11 books and V = 12.
Exercise 3: Show the worked answer
1. x = 72 ÷ 6 = 12 °C.
| xi | xi − x | (xi − x)² |
|---|---|---|
| 8 | −4 | 16 |
| 17 | 5 | 25 |
| 9 | −3 | 9 |
| 13 | 1 | 1 |
| 11 | −1 | 1 |
| 14 | 2 | 4 |
| Total | 0 | 56 |
2. V = 56 ÷ 6 ≈ 9.33.
3. σ = √V ≈ 3.06 °C (rounded to 2 decimal places).
b) V averages squared deviations, so its unit is °C². Taking the square root brings σ back to the unit of the data, °C, which is why σ is the figure we interpret.
Answer: V ≈ 9.33 °C² and σ ≈ 3.06 °C.
Exercise 4: Show the worked answer
1. x = 41 ÷ 5 = 8.2.
2. Squares of the values: 36, 100, 9, 144, 100. They add up to 389, so the mean of the squares is 389 ÷ 5 = 77.8.
3. x² = 8.2² = 67.24.
4. V = 77.8 − 67.24 = 10.56.
5. σ = √V ≈ 3.25 thousand steps.
This route skips the deviation column and gives exactly the same variance as the table method. Keep x exact (8.2 here): rounding the mean before squaring it throws the answer off.
Answer: V = 10.56 and σ ≈ 3.25.
Exercise 5: Show the worked answer
1. x = 1092 ÷ 6 = 182 cm.
| xi | xi − x | (xi − x)² |
|---|---|---|
| 172 | −10 | 100 |
| 195 | 13 | 169 |
| 184 | 2 | 4 |
| 171 | −11 | 121 |
| 190 | 8 | 64 |
| 180 | −2 | 4 |
| Total | 0 | 462 |
2. The squared deviations add up to 462.
a) σ² = 462 ÷ 6 = 77, so σ ≈ 8.77 cm.
b) s² = 462 ÷ 5 = 92.4, so s ≈ 9.61 cm.
c) The six players are only part of the club: they form a sample. To estimate the spread for the whole club, use s (divide by N − 1). The deviations are measured from the sample mean, which sits closer to the six values than the club's unknown mean does; dividing by N − 1 makes up for that. As a result, s is always slightly larger than σ.
Most school calculators show both values next to each other: the one that divides by N is often labelled σx, the one that divides by N − 1 is often labelled sx or Sx.
Answer: σ ≈ 8.77 cm; s ≈ 9.61 cm; use s for the club.
Exercise 6: Show the worked answer
Each value xi occurs ni times, so every value is weighted by its frequency.
| xi | ni | ni × xi | ni × xi² |
|---|---|---|---|
| 0 | 8 | 0 | 0 |
| 1 | 1 | 1 | 1 |
| 2 | 4 | 8 | 16 |
| 3 | 1 | 3 | 9 |
| 4 | 7 | 28 | 112 |
| 5 | 4 | 20 | 100 |
| Total | 25 | 60 | 238 |
1. N = 8 + 1 + 4 + 1 + 7 + 4 = 25.
2. x = Σ nixi ÷ N = 60 ÷ 25 = 2.4 books per student.
3. V = Σ nixi² ÷ N − x² = 238 ÷ 25 − 2.4² = 9.52 − 5.76 = 3.76.
4. σ = √V ≈ 1.94 books.
Answer: N = 25; x = 2.4; V = 3.76; σ ≈ 1.94.
Exercise 7: Show the worked answer
a) We do not know each student's exact time, only the class it falls in. We assume the students in a class are spread around its midpoint ci = (lower bound + upper bound) ÷ 2. Watch the last class, which is wider: its midpoint is 50.
| Class | ci | ni | ni × ci | ni × ci² |
|---|---|---|---|---|
| [0, 10) | 5 | 12 | 60 | 300 |
| [10, 20) | 15 | 7 | 105 | 1575 |
| [20, 30) | 25 | 5 | 125 | 3125 |
| [30, 40) | 35 | 8 | 280 | 9800 |
| [40, 60) | 50 | 8 | 400 | 20000 |
| Total | 40 | 970 | 34800 |
b) x ≈ 970 ÷ 40 = 24.25 min.
V ≈ 34800 ÷ 40 − 24.25² = 870 − 588.0625 = 281.9375.
σ ≈ √281.9375 ≈ 16.79 min.
These are estimates: they depend on the midpoint assumption, even though the arithmetic is exact.
Answer: x ≈ 24.25 min and σ ≈ 16.79 min.
Exercise 8: Show the worked answer
a) Group A: 96 ÷ 8 = 12. Group B: 96 ÷ 8 = 12. The means are equal.
b) Group A: squared deviations add up to 10, VA = 10 ÷ 8 = 1.25, σA ≈ 1.12.
Group B: squared deviations add up to 126, VB = 126 ÷ 8 = 15.75, σB ≈ 3.97.
c) σA < σB: group A's marks stay close to 12, while group B's are further away (range 4 points for A, 10 points for B). The mean alone cannot tell the two groups apart; the standard deviation can.
Answer: σA ≈ 1.12 and σB ≈ 3.97; group A is more consistent.
Exercise 9: Show the worked answer
a) x = 55 ÷ 5 = 11 €.
| xi | xi − x | (xi − x)² |
|---|---|---|
| 14 | 3 | 9 |
| 13 | 2 | 4 |
| 6 | −5 | 25 |
| 10 | −1 | 1 |
| 12 | 1 | 1 |
| Total | 0 | 40 |
V = 40 ÷ 5 = 8 and σ ≈ 2.83 €.
b) New prices: 16, 15, 8, 12, 14. Every price, and so the mean, goes up by €2: the mean becomes 13 €. The deviations xi − x are unchanged, so the standard deviation stays at σ ≈ 2.83 €.
c) A 10% rise means multiplying by 1.1. New prices: 15.4, 14.3, 6.6, 11, 13.2. The mean is multiplied by 1.1: 11 × 1.1 = 12.1 €. Each deviation is multiplied by 1.1 too, the variance by 1.1² = 1.21 and the standard deviation by 1.1: σ′ = 1.1 × σ ≈ 3.11 €.
Direct check on the new prices: V′ = 9.68 and √V′ ≈ 3.11, the same result.
Answer: a) x = 11 €, σ ≈ 2.83 €; b) 13 € and σ unchanged; c) 12.1 € and σ′ ≈ 3.11 €.
Exercise 10: Show the worked answer
a) x = 3233 ÷ 20 = 161.65 g. V = 527397 ÷ 20 − 161.65² = 26369.85 − 26130.7225 = 239.1275, so σ ≈ 15.46 g.
b) Use the unrounded σ for the bounds: x − 2σ ≈ 130.72 g and x + 2σ ≈ 192.58 g.
c) 19 apples out of 20 are inside the interval, which is 19 ÷ 20 × 100 = 95%. The only apple outside weighs 215 g: it is much heavier than the rest. A value this far from the mean is called an outlier; on its own it pushes up the standard deviation of the crate.
Answer: x = 161.65 g; σ ≈ 15.46 g; interval [130.72, 192.58]; 19 apples, i.e. 95%.
Answers at a glance
| Exercise | Results |
|---|---|
| 1. Finding a mean | x = 12 |
| 2. Variance of a small data set | x = 11; V = 12 |
| 3. Standard deviation and units | V ≈ 9.33; σ ≈ 3.06 °C |
| 4. The shortcut formula (mean of squares) | V = 10.56; σ ≈ 3.25 |
| 5. Population or sample: divide by N or by N − 1 | σ ≈ 8.77; s ≈ 9.61 |
| 6. Grouped data: frequency table | x = 2.4; V = 3.76; σ ≈ 1.94 |
| 7. Data grouped into classes | x ≈ 24.25; σ ≈ 16.79 |
| 8. Interpreting: same mean, different spread | σA ≈ 1.12; σB ≈ 3.97 |
| 9. What happens when every value changes | σ ≈ 2.83; +€2: σ unchanged; +10%: σ′ ≈ 3.11 |
| 10. The “mean ± 2 standard deviations” rule | x = 161.65; σ ≈ 15.46; 19/20 = 95% |
To check a result on your own data, use the site’s standard deviation calculator (in French): it gives the mean, the variance and the population standard deviation (÷N). See also the variance calculator and the mean calculator (in French).
How this worksheet is made: the data sets are drawn by a program from a fixed seed, and every mean, variance, standard deviation and worked step is recomputed by that program with exact fractions. No value is typed in by hand.
This worksheet is also available in French: Version française.