standarddeviationcalculator.net

Updated Free · runs in your browser

Learn

Standard deviation worksheets

Four printable standard deviation worksheets, from a first five-number data set up to grouped data and word problems, each with a full answer key and the working written out. Below them, a generator makes a new sheet of practice problems whenever you need one.

How to use these worksheets

The sheets are written for secondary school, AP and introductory college statistics, and for anyone relearning the method. Work through them in order: worksheet 1 is the calculation on its own, worksheet 2 adds the choice between sample and population and what the result means, worksheet 3 moves to frequency tables and grouped data, and worksheet 4 is about interpreting a standard deviation rather than computing one.

Print with your browser's print command or the button below. The site header, menus, the generator and the FAQ are left off the printout. Answer keys print only if you open them first, so a closed page gives a clean student copy and an opened one gives a teacher copy. Round final answers to 2 decimal places.

Make a new worksheet

Choose a difficulty and how many problems you want. Each click draws new data; the answer key, with the mean, the sum of squared deviations Σ(x − x̄)² and the standard deviation, is under the working panel.

From 3 to 10.

Worksheet 1: population and sample standard deviation

Show the mean, the deviations and the sum of squares for each problem. Round final answers to 2 decimal places.

  1. The five players on a team scored 2, 4, 6, 8 and 10 points. Treating the team as the whole population, find the variance and standard deviation.
  2. A sample of four test tubes holds 3, 5, 7 and 9 ml. Find the sample variance and sample standard deviation.
  3. A data set is 5, 5, 5, 5, 5. Without a long calculation, give its sample and population standard deviations, and say why.
  4. For the data 1, 3, 4, 6, 8, 8, find both the sample and the population standard deviation.
  5. A sample of five commutes took 12, 15, 11, 18 and 14 minutes. Find the sample standard deviation.
  6. A shop sold 7, 2, 9, 4, 6, 3 and 11 umbrellas on the seven days of one week, and only that week is of interest. Find the population standard deviation.
  7. For the data 20, 22, 25, 19, 24, 26, find both standard deviations. (The mean is not a whole number; the shortcut Σx² − (Σx)² ÷ n helps.)
Show answers and working: worksheet 1

  1. Mean = 30 ÷ 5 = 6. Deviations −4, −2, 0, 2, 4; squares 16, 4, 0, 4, 16; Σ(x − x̄)² = 40.

    σ² = 40 ÷ 5 = 8 σ = √8 = 2.83

  2. Mean = 24 ÷ 4 = 6. Deviations −3, −1, 1, 3; squares 9, 1, 1, 9; Σ(x − x̄)² = 20.

    s² = 20 ÷ 3 = 6.67 s = √6.667 = 2.58

  3. Both are 0. Every value equals the mean of 5, so every deviation is 0 and the sum of squares is 0, whichever divisor you use.

  4. Mean = 30 ÷ 6 = 5. Deviations −4, −2, −1, 1, 3, 3; squares 16, 4, 1, 1, 9, 9; Σ(x − x̄)² = 40.

    s = √(40 ÷ 5) = √8 = 2.83 σ = √(40 ÷ 6) = √6.667 = 2.58

  5. Mean = 70 ÷ 5 = 14. Deviations −2, 1, −3, 4, 0; squares 4, 1, 9, 16, 0; Σ(x − x̄)² = 30.

    s² = 30 ÷ 4 = 7.5 s = √7.5 = 2.74 minutes

  6. Mean = 42 ÷ 7 = 6. Deviations 1, −4, 3, −2, 0, −3, 5; squares 1, 16, 9, 4, 0, 9, 25; Σ(x − x̄)² = 64.

    σ² = 64 ÷ 7 = 9.14 σ = √9.143 = 3.02 umbrellas

  7. Σx = 136, so the mean is 136 ÷ 6 = 22.67 (22.6667). Σx² = 400 + 484 + 625 + 361 + 576 + 676 = 3,122.

    Σ(x − x̄)² = 3,122 − 136² ÷ 6 = 3,122 − 3,082.667 = 39.333 s = √(39.333 ÷ 5) = √7.867 = 2.80 σ = √(39.333 ÷ 6) = √6.556 = 2.56

Worksheet 2: sample or population, variance, mean ± SD and CV

Round final answers to 2 decimal places.

  1. A tutor records the quiz scores of all six students in her tutorial group, 8, 9, 6, 7, 10, 8, and wants to describe only this group. Should she use the sample or the population standard deviation? Calculate it.
  2. (a) The variance of a data set is 49. What is the standard deviation? (b) A standard deviation is 3.5. What is the variance? (c) A variance is 18. What is the standard deviation?
  3. An inspector measures eight bolts picked at random from a large batch: 14, 16, 15, 13, 17, 15, 16, 14 mm. Find the sample standard deviation and write the result in the form mean ± SD.
  4. Test scores have a mean of 72 and a standard deviation of 6. Give the interval of scores within 1 standard deviation of the mean, and within 2 standard deviations.
  5. Five sample packets weigh 40, 44, 38, 42 and 46 g. Find the sample standard deviation and the coefficient of variation (CV = s ÷ x̄, as a percentage).
  6. In one group, heights have a mean of 150 cm with a standard deviation of 6 cm; weights have a mean of 30 kg with a standard deviation of 3 kg. Which varies more relative to its mean? Why can you not simply compare 6 with 3?
  7. A student reports a population standard deviation of 4 for five values, but the values were a sample. Without the raw data, find the correct sample standard deviation.
Show answers and working: worksheet 2

  1. Population: the six students are the whole group she is describing. Mean = 48 ÷ 6 = 8. Deviations 0, 1, −2, −1, 2, 0; Σ(x − x̄)² = 10.

    σ = √(10 ÷ 6) = √1.667 = 1.29

    (Using n − 1 by mistake would give √2 = 1.41.)

  2. (a) √49 = 7. (b) 3.5² = 12.25. (c) √18 = 4.24.

  3. Sample, because the bolts stand for the whole batch. Mean = 120 ÷ 8 = 15. Deviations −1, 1, 0, −2, 2, 0, 1, −1; squares sum to 12.

    s² = 12 ÷ 7 = 1.714 s = √1.714 = 1.31 mm Result: 15 ± 1.31 mm (13.69 to 16.31 mm)

  4. Within 1 SD: 72 − 6 to 72 + 6 = 66 to 78. Within 2 SD: 72 − 12 to 72 + 12 = 60 to 84.

  5. Mean = 210 ÷ 5 = 42. Deviations −2, 2, −4, 0, 4; squares 4, 4, 16, 0, 16; Σ(x − x̄)² = 40.

    s = √(40 ÷ 4) = √10 = 3.16 g CV = 3.162 ÷ 42 = 0.0753 = 7.53%

  6. CV(height) = 6 ÷ 150 = 4% CV(weight) = 3 ÷ 30 = 10%

    Weight varies more relative to its mean. The raw standard deviations are in different units (cm and kg), so 6 and 3 cannot be compared; the CV has no units.

  7. Recover the sum of squares from the population formula, then divide by n − 1 instead.

    Σ(x − x̄)² = n × σ² = 5 × 16 = 80 s = √(80 ÷ 4) = √20 = 4.47

    In general s = σ × √(n ÷ (n − 1)) = 4 × √1.25 = 4.47.

Worksheet 3: frequency tables, grouped data and z-scores

For grouped data, use the midpoint of each class. Round final answers to 2 decimal places.

  1. A sample of 20 students were asked how many books they read last month. Find the mean, the sample standard deviation and the population standard deviation.
    Books (x)12345
    Students (f)25841
  2. A team's goals in each of the 20 matches of its season (the whole season is the population of interest). Find the mean and the population standard deviation.
    Goals (x)01234
    Matches (f)58421
  3. All 30 workers in a workshop were timed on a task. Estimate the mean and the population standard deviation.
    Minutes0–1010–2020–3030–4040–50
    Workers (f)371262
  4. The heights of a random sample of 40 students are grouped below. Estimate the mean and the sample standard deviation.
    Height (cm)150–155155–160160–165165–170170–175
    Students (f)491485
  5. An exam has a mean of 68 and a standard deviation of 8. Find the z-scores of marks of 80 and 60, and say what each means.
  6. Using the mean and population standard deviation from problem 1, find the z-scores of a student who read 5 books and one who read 1 book. Which is further from typical?
Show answers and working: worksheet 3

  1. n = Σf = 20. Σfx = 2 + 10 + 24 + 16 + 5 = 57, so the mean is 57 ÷ 20 = 2.85 books. Σfx² = 2 + 20 + 72 + 64 + 25 = 183.

    Σf(x − x̄)² = 183 − 57² ÷ 20 = 183 − 162.45 = 20.55 s = √(20.55 ÷ 19) = √1.0816 = 1.04 σ = √(20.55 ÷ 20) = √1.0275 = 1.01

  2. Σfx = 0 + 8 + 8 + 6 + 4 = 26, so the mean is 26 ÷ 20 = 1.3 goals. Σfx² = 0 + 8 + 16 + 18 + 16 = 58.

    Σf(x − x̄)² = 58 − 26² ÷ 20 = 58 − 33.8 = 24.2 σ² = 24.2 ÷ 20 = 1.21 σ = √1.21 = 1.10 goals

  3. Midpoints 5, 15, 25, 35, 45. Σfm = 15 + 105 + 300 + 210 + 90 = 720, so the mean is 720 ÷ 30 = 24 minutes.

    f(m − x̄)²: 3 × 361 + 7 × 81 + 12 × 1 + 6 × 121 + 2 × 441 = 1,083 + 567 + 12 + 726 + 882 = 3,270 σ² = 3,270 ÷ 30 = 109 σ = √109 = 10.44 minutes

    This is an estimate: the midpoint stands in for every value in its class.

  4. Midpoints 152.5, 157.5, 162.5, 167.5, 172.5. Σfm = 610 + 1,417.5 + 2,275 + 1,340 + 862.5 = 6,505, so the mean is 6,505 ÷ 40 = 162.63 cm (162.625).

    f(m − x̄)²: 4 × 102.5156 + 9 × 26.2656 + 14 × 0.0156 + 8 × 23.7656 + 5 × 97.5156 = 410.0625 + 236.3906 + 0.2188 + 190.125 + 487.5781 = 1,324.375 s² = 1,324.375 ÷ 39 = 33.96 s = √33.958 = 5.83 cm

  5. z(80) = (80 − 68) ÷ 8 = 1.5 z(60) = (60 − 68) ÷ 8 = −1

    80 is one and a half standard deviations above the mean; 60 is one standard deviation below it.

  6. From problem 1: mean 2.85, σ = 1.0137.

    z(5) = (5 − 2.85) ÷ 1.0137 = 2.12 z(1) = (1 − 2.85) ÷ 1.0137 = −1.83

    The student who read 5 books is further from typical: |2.12| is larger than |−1.83|.

Worksheet 4: word problems and interpretation

Use the sample standard deviation unless told otherwise. Round final answers to 2 decimal places.

  1. Two machines fill 500 g bags. Five bags from machine A weigh 498, 502, 500, 499, 501 g; five from machine B weigh 495, 505, 497, 503, 500 g. Find each mean and standard deviation. Which machine is more consistent?
  2. Find the standard deviation of 4, 7, 9, 10, 15. Then 5 is added to every value. Without recalculating from scratch, give the new mean and standard deviation.
  3. Every value in 4, 7, 9, 10, 15 is instead multiplied by 3. Give the new mean, standard deviation and variance.
  4. A sample of six reaction times is 10, 12, 11, 13, 12, 11 hundredths of a second. A seventh reading of 30 is then added. Find the mean, median and standard deviation before and after. Which measure changed most?
  5. Adult heights in a population are bell-shaped with a mean of 170 cm and a standard deviation of 7 cm. Using the empirical rule, give the ranges that hold about 68%, 95% and 99.7% of heights, and the percentage taller than 184 cm.
  6. A machine makes 400 bolts with a mean length of 20.0 mm and a standard deviation of 0.1 mm, bell-shaped. About how many bolts are between 19.8 and 20.2 mm? About how many are longer than 20.2 mm?
  7. Five noon temperatures were 18.0, 21.5, 19.0, 24.0 and 22.5 °C. Find the mean and standard deviation. Then convert to Fahrenheit (F = 1.8C + 32) and give the new mean and standard deviation.
Show answers and working: worksheet 4

  1. Both means are 2,500 ÷ 5 = 500 g.

    A: deviations −2, 2, 0, −1, 1; Σ = 10; s = √(10 ÷ 4) = 1.58 g B: deviations −5, 5, −3, 3, 0; Σ = 68; s = √(68 ÷ 4) = 4.12 g

    Machine A is more consistent: same average, much smaller spread.

  2. Mean = 45 ÷ 5 = 9. Deviations −5, −2, 0, 1, 6; squares 25, 4, 0, 1, 36; Σ(x − x̄)² = 66.

    s² = 66 ÷ 4 = 16.5 s = √16.5 = 4.06

    After adding 5: mean = 9 + 5 = 14, standard deviation still 4.06. Every value and the mean move together, so no deviation changes.

  3. The new data are 12, 21, 27, 30, 45. Every deviation is tripled, so every squared deviation is multiplied by 9.

    mean = 3 × 9 = 27 s = 3 × 4.062 = 12.19 variance = 9 × 16.5 = 148.5

  4. Before: mean = 69 ÷ 6 = 11.5, median = 11.5, Σ(x − x̄)² = 5.5, s = √(5.5 ÷ 5) = 1.05.

    After: mean = 99 ÷ 7 = 14.14, median = 12, Σ(x − x̄)² = 298.86, s = √(298.86 ÷ 6) = 7.06.

    The standard deviation changed most: it grew almost sevenfold, the mean rose by 2.64 and the median by only 0.5. Squaring the outlier's deviation of 15.86 gives it most of the sum of squares.

  5. 68%: 170 ± 7 → 163 to 177 cm 95%: 170 ± 14 → 156 to 184 cm 99.7%: 170 ± 21 → 149 to 191 cm

    184 cm is 2 standard deviations above the mean. About 5% lie outside ±2 SD, split evenly between the two tails, so about 2.5% are taller than 184 cm.

  6. 19.8 and 20.2 mm are 20.0 ∓ 2 × 0.1, so the interval is ±2 standard deviations and holds about 95%: 0.95 × 400 = about 380 bolts. Longer than 20.2 mm is the upper tail, about 2.5%: 0.025 × 400 = about 10 bolts.

  7. Mean = 105 ÷ 5 = 21 °C. Deviations −3, 0.5, −2, 3, 1.5; squares 9, 0.25, 4, 9, 2.25; Σ(x − x̄)² = 24.5.

    s = √(24.5 ÷ 4) = √6.125 = 2.47 °C Fahrenheit mean = 1.8 × 21 + 32 = 69.8 °F Fahrenheit SD = 1.8 × 2.4749 = 4.45 °F

    The +32 shifts every value equally and has no effect on the spread; only the × 1.8 changes the standard deviation.

Checking your work and further practice

If a step in the working is unclear, the guide on how to calculate standard deviation goes through the method one line at a time. The sample vs population page explains the n − 1 choice behind worksheet 2. To check any answer, paste the data into the standard deviation calculator, which shows the same working. For timed practice with instant marking, try the standard deviation quiz or the online practice problems, which mark each calculation and show the working.

Standard deviation worksheets: the worked example on this page, with its result and chart
Standard deviation worksheets: the worked example above, at a glance.

Common questions

Do the answers print with the worksheets?

Only if you open them. Each answer key sits in a closed "Show answers and working" panel under its worksheet, and a closed panel is left out of the printout. Open the panels before printing when you want a teacher copy with the full working.

What rounding do the answer keys use?

Final answers are rounded to 2 decimal places. Intermediate values such as the mean are kept unrounded in the working (shown to 2–4 decimal places where they do not come out exactly), because rounding the mean first can change the last digit of the standard deviation.

Should students use the sample or the population formula?

Each problem says which one, or asks the student to decide. The rule: if the data are every member of the group you care about, divide by n (population, σ); if they are a selection used to describe a larger group, divide by n − 1 (sample, s). Worksheet 2 practises exactly this choice.

How is the generator different from the four fixed worksheets?

The fixed worksheets cover several skills: frequency tables, grouped data, z-scores and interpretation. The generator practises the core calculation only, but with fresh numbers every time, so a student can keep going until the method is automatic. Its answer key gives the mean, the sum of squared deviations and the standard deviation for each set.

Can students check their own answers with a calculator?

Yes. Paste any data set into the standard deviation calculator and it shows the mean, both standard deviations and every step of the working. For worksheet 3, the frequency table and grouped data calculators take the tables directly.

Written and reviewed by our editorial team. Last updated . Method and sources: how these numbers are computed.