Calculate the standard deviation from the formula

Times in minutes

\(\displaystyle 27,\ 27,\ 30,\ 33,\ 33\)

Mean time

\(\displaystyle 30\text{ minutes}\)

Five deliveries took 27, 27, 30, 33 and 33 minutes. The mean time was 30 minutes. What is the sample standard deviation of the five times?

Five plants have heights that spread out as the slider moves. At 0 all five plants are the same height

Move the Spread slider through every position. As the plants spread out, what stayed the same and what changed?

Calculating standard deviation

For the sample standard deviation, square each deviation from the exact mean, add, divide by n−1 and take the non-negative square root. Use at least two observations

1. 1Find the mean Add all values and divide by the number of values \(\displaystyle \bar{x}=\frac{27+27+30+33+33}{5}=\frac{150}{5}=30\)
2. 2Use signed deviations Subtract the mean from each value; values below the mean have negative deviations \(\displaystyle \left(27-30,27-30,30-30,33-30,33-30\right)=\left(-3,-3,0,3,3\right)\)
3. 3Square before adding Square every deviation so negative and positive deviations cannot cancel \(\displaystyle \left(-3\right)^{2}+\left(-3\right)^{2}+0^{2}+3^{2}+3^{2}=9+9+0+9+9=36\)
4. 4Divide and take the root Divide the sum of the squared deviations by n - 1, then take the non-negative square root \(\displaystyle s=\sqrt{\frac{36}{5-1}}=\sqrt{9}=3\)

Choose the stated standard deviation

These questions ask for the sample standard deviation s, using divisor n−1 and at least two observations. A population standard deviation uses divisor n. They answer different specifications; a calculator may display both

1. 1Interpret units Variance is measured in squared units; standard deviation is in the original measurement units. Standard deviation is not a frequency, and it need not equal an observed deviation
2. 2Zero and undefined For n≥2, sample standard deviation is zero exactly when all observations agree. For n=1, the sample formula has divisor zero and is undefined; do not report zero as its sample standard deviation
3. 3Sensitivity and limits Every squared deviation contributes. An extreme observation can strongly affect s. No fixed proportion within one standard deviation follows for arbitrary data; a normal-model rule requires its own assumptions
4. 4Keep precision Use the unrounded mean and squared-deviation total, then round only the requested final result. Report the requested number of decimal places

A sample contains one recorded value, 12 cm. What is its sample standard deviation using divisor n−1?

Count and squared-deviation sum

\(\displaystyle n=6,\quad SS=60\text{ cm}^2\)

Six heights have squared deviations totalling 60 cm². What is the sample variance, in cm²?

Tray 1 heights in centimetres

\(\displaystyle 11,\ 14,\ 15,\ 17,\ 18,\ 21\)

Tray 2 heights in centimetres

\(\displaystyle 13,\ 15,\ 16,\ 16,\ 17,\ 19\)

Two trays of six seedlings were measured. Find the sample standard deviation of the heights in tray 1, to 1 decimal place

Find the mean of the six heights first, then work out the six deviations from it

Tray 1 heights in centimetres

\(\displaystyle 11,\ 14,\ 15,\ 17,\ 18,\ 21\)

Tray 2 heights in centimetres

\(\displaystyle 13,\ 15,\ 16,\ 16,\ 17,\ 19\)

Using the same two trays, find the sample standard deviation of the heights in tray 2.

Find the mean of the six heights first, then square each deviation before adding

Times in minutes

\(\displaystyle 27,\ 27,\ 30,\ 33,\ 33\)

Squared deviations added

\(\displaystyle 36\)

Omitting the square root. One of these lines is right. Which one?

Heights (cm)

\(\displaystyle 16,\ 16,\ 16,\ 16,\ 16,\ 16\)

Six recorded heights are all 16 cm. Find their sample standard deviation in cm

For arbitrary data, does a standard deviation of 3 cm mean every observation is within 3 cm of the mean?

Compare complete sample-spread calculations

For tray A: 11, 14, 15, 17, 18, 21 cm and tray B: 13, 15, 16, 16, 17, 19 cm, produce complete tables of values, deviations and squared deviations. Calculate count, mean, sample variance, sample standard deviation, range and IQR. Use medians of halves for quartiles. Give each standard deviation to one decimal place and compare the trays using numbers and units

1. 1Choose and use technology Choose statistical functions or spreadsheet formulas, explain the choice and record raw inputs, the sample/population setting, outputs and final rounding. Identify both sample and population standard-deviation outputs for A and explain which the task requires
2. 2Verify independently Manually check each deviation, the signed sum, squared sum and divisor. Compare your result with the chosen function, then check the comparison using range and IQR. A zero signed sum alone does not prove the squared-deviation calculation is correct
3. 3Change location and units Predict and check the effect of adding 5 cm to every A height. Separately convert every B height from centimetres to millimetres. Explain which quantities change, and why spread values in different units cannot be compared directly
4. 4Complete before continuing Written work and external-tool execution are not automatically marked; use self or teacher review

Model response: sample spread and independent checks

Tray A heights are 11, 14, 15, 17, 18 and 21 cm; tray B heights are 13, 15, 16, 16, 17 and 19 cm. Both trays have six observations, mean and median 16 cm. A has the larger sample standard deviation, approximately 3.5 cm versus 2.0 cm; its range and IQR are also larger. This describes these recorded trays without establishing a cause or a population difference

1. 1Full calculation for A Deviations from 16 are −5, −2, −1, 1, 2, 5 cm. Their sum is zero; squared deviations are 25, 4, 1, 1, 4, 25 cm² and total 60 cm². Sample variance is 60/5=12 cm² and s=√12 cm, approximately 3.5 cm to one decimal place. Range is 21−11=10 cm; Q1=14 and Q3=18 give IQR 4 cm
2. 2Full calculation for B Deviations are −3, −1, 0, 0, 1, 3 cm, summing to zero. Squared deviations 9, 1, 0, 0, 1, 9 total 20 cm². Sample variance is 20/5=4 cm² and s=2 cm, written 2.0 cm to one decimal place. Range is 19−13=6 cm; Q1=15 and Q3=17 give IQR 2 cm
3. 3One reproducible workflow Enter A in A2:A7. Use COUNT(A2:A7), AVERAGE(A2:A7), VAR.S(A2:A7) and STDEV.S(A2:A7). For an independent formula table, put the mean in E2, set B2=A2−$E$2 and C2=B2^2, and fill down. Check SUM(B2:B7)=0, SUM(C2:C7)=60 and SQRT(SUM(C2:C7)/(COUNT(A2:A7)−1)) agrees with STDEV.S. Repeat in separate columns for B. Record full precision internally and format final s to one decimal place
4. 4Different divisor For A, STDEV.P would use 60/6=10 under the root, giving approximately 3.2 cm to one decimal place. It does not answer this sample-standard-deviation request. For B, the corresponding population value is √(20/6), approximately 1.8 cm. Tool labels vary, so verify the displayed definition instead of relying on position in a results list
5. 5Shift and conversion checks Adding 5 cm to every A height raises its mean and median to 21 cm, but leaves every deviation, range, IQR, variance and standard deviation unchanged. Converting B to millimetres multiplies every value, standard deviation, range and IQR by 10: mean 160 mm, s=20 mm, range 60 mm and IQR 20 mm. Variance becomes 400 mm², multiplying by 100. Convert to common units before comparing
6. 6Review the whole process Check every table entry, count, exact mean, two sums, divisor, square root, units, requested precision, comparison and transformation. Retain unusual records unless there is a stated defensible reason to change them. The model supports self or teacher review; it does not certify independent tool operation

Tray B has a sample standard deviation of 2 cm and sample variance of 4 cm². All its heights are converted from cm to mm. Which statement is correct?