Standard deviation and variance practice problems

By Jude Wallis · Published July 28, 2026

This set has eight practice problems on standard deviation and variance, from routine by-hand calculations to AP-style multi-part reasoning. Work each one first, then open the solution to check every arithmetic move against the mean, the squared deviations, and the divisor.

AP Statistics: Unit 1 (topics 1.7 Summary Statistics for One Quantitative Variable). Standard deviation and variance are Unit 1 topic 1.7 (learning objective 1.7.B); the effect of changing units of measurement on summary statistics is 1.7.C in the Fall 2026 AP Statistics course.

What these problems build

These eight problems build fluency with the by-hand standard deviation routine: find the mean, square each deviation, divide by $n-1$ for a sample or $n$ for a population, then take the square root. The later problems add a second skill, predicting how the mean, variance, and standard deviation change when you add a constant to every value or multiply every value by a constant. If you want the method laid out step by step first, read standard deviation by hand, and use the standard deviation calculator to confirm your answers. For more sets like this one, see all practice problems.

Problem 1

A wildlife trail camera ran for five nights and photographed 5, 7, 8, 9, and 11 foxes. Treat these five nights as the entire period of interest, so the data is a population. Find the population variance and the population standard deviation.

Show the worked solution

Find the mean $\mu$ (say "mu"). Add the values and divide by $n = 5$ : $\mu = \frac{5 + 7 + 8 + 9 + 11}{5} = \frac{40}{5} = 8$ foxes.
Subtract the mean from each value to get the deviations $x_i - \mu$ : $5 - 8 = -3$ , $7 - 8 = -1$ , $8 - 8 = 0$ , $9 - 8 = 1$ , $11 - 8 = 3$ . They sum to $-3 - 1 + 0 + 1 + 3 = 0$ , a good check.
Square each deviation: $(-3)^2 = 9$ , $(-1)^2 = 1$ , $0^2 = 0$ , $1^2 = 1$ , $3^2 = 9$ .
Add the squared deviations: $9 + 1 + 0 + 1 + 9 = 20$ .
Divide by $n = 5$ for the population variance: $\sigma^2 = \frac{20}{5} = 4$ .
Take the square root for the population standard deviation: $\sigma = \sqrt{4} = 2$ foxes.

The population variance is $\sigma^2 = 4$ (foxes squared) and the population standard deviation is $\sigma = 2$ foxes.

Problem 2

A commuter records how many minutes her train arrives late on 5 randomly chosen weekdays: 6, 8, 10, 12, and 14 minutes. Treat these days as a sample. Find the sample variance and the sample standard deviation.

Show the worked solution

Find the mean $\bar{x}$ (say "x-bar"). Add the values and divide by $n = 5$ : $\bar{x} = \frac{6 + 8 + 10 + 12 + 14}{5} = \frac{50}{5} = 10$ minutes.
Subtract the mean from each value: $6 - 10 = -4$ , $8 - 10 = -2$ , $10 - 10 = 0$ , $12 - 10 = 2$ , $14 - 10 = 4$ . The deviations sum to $0$ .
Square each deviation: $(-4)^2 = 16$ , $(-2)^2 = 4$ , $0^2 = 0$ , $2^2 = 4$ , $4^2 = 16$ .
Add the squared deviations: $16 + 4 + 0 + 4 + 16 = 40$ .
Divide by $n - 1 = 4$ for the sample variance: $s^2 = \frac{40}{4} = 10$ .
Take the square root for the sample standard deviation: $s = \sqrt{10} \approx 3.16$ minutes.

The sample variance is $s^2 = 10$ (minutes squared) and the sample standard deviation is $s = \sqrt{10} \approx 3.16$ minutes.

Problem 3

A food scientist measures the sugar content of 5 energy bars pulled from one production line: 13, 19, 20, 21, and 27 grams. Find the sample standard deviation, then the population standard deviation, and explain which is larger and why.

Show the worked solution

Find the mean: $\bar{x} = \frac{13 + 19 + 20 + 21 + 27}{5} = \frac{100}{5} = 20$ grams.
Deviations from the mean: $13 - 20 = -7$ , $19 - 20 = -1$ , $20 - 20 = 0$ , $21 - 20 = 1$ , $27 - 20 = 7$ .
Square them: $(-7)^2 = 49$ , $(-1)^2 = 1$ , $0^2 = 0$ , $1^2 = 1$ , $7^2 = 49$ , and add: $49 + 1 + 0 + 1 + 49 = 100$ .
The sample version divides by $n - 1 = 4$ : $s^2 = \frac{100}{4} = 25$ , so $s = \sqrt{25} = 5$ grams.
The population version divides by $n = 5$ : $\sigma^2 = \frac{100}{5} = 20$ , so $\sigma = \sqrt{20} \approx 4.47$ grams.
The sample value is larger because it divides the same sum of squares, $100$ , by the smaller number $4$ instead of $5$ .

The sample standard deviation is $s = \sqrt{25} = 5$ grams and the population standard deviation is $\sigma = \sqrt{20} \approx 4.47$ grams; the sample value is larger because it divides by $n - 1 = 4$ rather than $n = 5$ .

Problem 4

A fitness tracker logs the number of flights of stairs a user climbs on 5 days: 21, 23, 24, 25, and 27 flights. The tracker was over-counting by 5 flights each day, so 5 is subtracted from every value. Treating the days as a sample, find the mean and standard deviation before and after the correction.

Show the worked solution

Original mean: $\bar{x} = \frac{21 + 23 + 24 + 25 + 27}{5} = \frac{120}{5} = 24$ flights.
Original deviations: $21 - 24 = -3$ , $23 - 24 = -1$ , $24 - 24 = 0$ , $25 - 24 = 1$ , $27 - 24 = 3$ . Squared they are $9, 1, 0, 1, 9$ , summing to $20$ .
Original sample standard deviation: $s = \sqrt{\frac{20}{5 - 1}} = \sqrt{\frac{20}{4}} = \sqrt{5} \approx 2.24$ flights.
Subtract 5 from each value to get $16, 18, 19, 20, 22$ . New mean: $\frac{16 + 18 + 19 + 20 + 22}{5} = \frac{95}{5} = 19$ flights, which is $24 - 5$ .
New deviations: $16 - 19 = -3$ , $18 - 19 = -1$ , $19 - 19 = 0$ , $20 - 19 = 1$ , $22 - 19 = 3$ . They match the originals, so the squared deviations again sum to $20$ .
New sample standard deviation: $s = \sqrt{\frac{20}{4}} = \sqrt{5} \approx 2.24$ flights, unchanged. Subtracting a constant shifts the mean by that constant but leaves the spread alone.

The mean drops by 5, from $24$ to $19$ flights, while the standard deviation stays $\sqrt{5} \approx 2.24$ flights; adding or subtracting a constant never changes the spread.

Problem 5

A juice bar triples a recipe. The amounts of its 5 ingredients are 2, 4, 6, 8, and 10 ounces, and every amount is multiplied by 3 for the triple batch. Treating the amounts as a sample, find the mean, variance, and standard deviation before and after tripling.

Show the worked solution

Original mean: $\bar{x} = \frac{2 + 4 + 6 + 8 + 10}{5} = \frac{30}{5} = 6$ ounces.
Original deviations: $-4, -2, 0, 2, 4$ ; squared they are $16, 4, 0, 4, 16$ , summing to $40$ .
Original sample variance and standard deviation: $s^2 = \frac{40}{5 - 1} = \frac{40}{4} = 10$ and $s = \sqrt{10} \approx 3.16$ ounces.
Multiply every value by 3 to get $6, 12, 18, 24, 30$ . New mean: $\frac{6 + 12 + 18 + 24 + 30}{5} = \frac{90}{5} = 18$ ounces, which is $6 \times 3$ .
New deviations: $-12, -6, 0, 6, 12$ ; squared they are $144, 36, 0, 36, 144$ , summing to $360$ . New variance: $s^2 = \frac{360}{4} = 90$ .
New standard deviation: $s = \sqrt{90} = 3\sqrt{10} \approx 9.49$ ounces. Multiplying by 3 multiplies the mean and standard deviation by 3 and the variance by $3^2 = 9$ .

After tripling, the mean is $18$ ounces, the variance is $90$ , and the standard deviation is $\sqrt{90} \approx 9.49$ ounces; multiplying by 3 scales the standard deviation by 3 and the variance by $9$ .

Problem 6

A weather log holds 8 daily high temperatures with a mean of 15 degrees Celsius and a standard deviation of 5 degrees Celsius. Each value is converted to Fahrenheit with $F = 1.8C + 32$ , where $C$ is the Celsius reading. Find the mean, standard deviation, and variance of the Fahrenheit values without recomputing from the raw data.

Show the worked solution

The conversion $F = 1.8C + 32$ multiplies every value by $1.8$ and then adds $32$ .
The mean follows the whole conversion: new mean $= 1.8 \times 15 + 32 = 27 + 32 = 59$ degrees Fahrenheit.
Adding $32$ shifts every value equally, so it does not change the spread. Only the multiplier $1.8$ affects the standard deviation.
New standard deviation $= 1.8 \times 5 = 9$ degrees Fahrenheit.
The variance scales by the square of the multiplier: the original variance is $5^2 = 25$ , so the new variance $= 1.8^2 \times 25 = 3.24 \times 25 = 81$ .
As a check, $\sqrt{81} = 9$ , which matches the standard deviation from step 4.

The Fahrenheit values have mean $59$ , standard deviation $9$ , and variance $81$ ; the $+32$ shifts the center only, while the $\times 1.8$ scales the spread.

Problem 7

Two checkout lanes each served customers over 5 one-hour blocks. Lane A served 8, 9, 10, 11, and 12 customers; Lane B served 4, 7, 10, 13, and 16 customers. Both lanes have the same mean. Decide which lane has the larger sample standard deviation, then compute both to confirm.

Show the worked solution

Both means are equal: Lane A gives $\frac{8 + 9 + 10 + 11 + 12}{5} = \frac{50}{5} = 10$ and Lane B gives $\frac{4 + 7 + 10 + 13 + 16}{5} = \frac{50}{5} = 10$ customers.
Lane B's values sit farther from $10$ than Lane A's, so Lane B should have the larger standard deviation.
Lane A deviations: $-2, -1, 0, 1, 2$ ; squared they are $4, 1, 0, 1, 4$ , summing to $10$ . Sample variance $= \frac{10}{4} = 2.5$ , so $s_A = \sqrt{2.5} \approx 1.58$ customers.
Lane B deviations: $-6, -3, 0, 3, 6$ ; squared they are $36, 9, 0, 9, 36$ , summing to $90$ . Sample variance $= \frac{90}{4} = 22.5$ , so $s_B = \sqrt{22.5} \approx 4.74$ customers.
The numbers confirm the prediction: $s_B \approx 4.74$ is much larger than $s_A \approx 1.58$ , even though both lanes average 10 customers.

Lane B has the larger spread: $s_A = \sqrt{2.5} \approx 1.58$ versus $s_B = \sqrt{22.5} \approx 4.74$ customers, while both means equal $10$ .

Problem 8

A coffee roaster weighs the beans lost during roasting for 6 batches: 14, 18, 20, 20, 22, and 26 grams, treated as a sample. (a) Find the sample mean, variance, and standard deviation. (b) The scale read 4 grams low, so 4 is added to every weight; give the new mean and standard deviation. (c) Those corrected weights are then multiplied by 0.5 to convert to an index; give the mean and standard deviation of the index.

Show the worked solution

Part (a) mean: $\bar{x} = \frac{14 + 18 + 20 + 20 + 22 + 26}{6} = \frac{120}{6} = 20$ grams.
Deviations: $-6, -2, 0, 0, 2, 6$ ; squared they are $36, 4, 0, 0, 4, 36$ , summing to $80$ .
Part (a) variance and standard deviation: $s^2 = \frac{80}{6 - 1} = \frac{80}{5} = 16$ and $s = \sqrt{16} = 4$ grams.
Part (b): adding 4 shifts the mean to $20 + 4 = 24$ grams and leaves the spread unchanged, so the standard deviation stays $4$ grams.
Part (c): multiplying by $0.5$ scales the mean to $24 \times 0.5 = 12$ and the standard deviation to $4 \times 0.5 = 2$ .
So the index has mean $12$ and standard deviation $2$ , and its variance is $2^2 = 4$ .

(a) mean $20$ grams, variance $16$ , standard deviation $4$ grams; (b) mean $24$ grams, standard deviation $4$ grams; (c) index mean $12$ , standard deviation $2$ .