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Expected value and variability · Tutorial 327 of 1000

The Standard Deviation of a Discrete Random Variable

Learn to turn a discrete random variable’s variance into standard deviation, using a column-by-column calculation and reporting the result in the original units.

Intermediate 9 min read

What You'll Learn

  • Calculate standard deviation as the square root of variance.
  • Organize deviations, squared deviations, and probability-weighted contributions in columns.
  • Check that a distribution is valid and calculate its mean before finding spread.
  • Report standard deviation in the same units as the random variable.
  • Distinguish standard deviation from variance and from the range of possible values.

From Variance to Standard Deviation

In Variance of a Discrete Random Variable, you found spread by calculating the probability-weighted average of squared distances from the mean. The result, \(\sigma_X^2\), is useful, but its units are squared. Standard deviation takes the square root of variance, giving a measure of spread in the same units as the random variable.

For a discrete random variable \(X\), calculate the variance first, using the mean \(\mu_X\) and the probabilities in its distribution. Then take the nonnegative square root. The standard deviation is written \(\sigma_X\). It is not a new calculation that replaces the variance table: it is the final step after the table’s weighted squared deviations have been added.

Definition: The standard deviation of a discrete random variable \(X\), written \(\sigma_X\), is the nonnegative square root of its variance \(\sigma_X^2\).
$$ \sigma_X=\sqrt{\sigma_X^2} =\sqrt{\sum (x-\mu_X)^2P(X=x)} $$

The square root is nonnegative because a standard deviation describes a distance-based measure of spread. A variance cannot be negative: every squared deviation is nonnegative, every probability is nonnegative, and the variance adds the products of those quantities.

Units provide a useful way to distinguish the two measures. If \(X\) measures minutes, \(\sigma_X^2\) is in minutes squared, while \(\sigma_X\) is in minutes. If \(X\) counts items, variance is in items squared and standard deviation is in items. Taking the square root restores the original units; it does not change the random variable or its distribution.

Organizing the Calculation by Columns

A clear table makes it easier to keep the variance calculation and the square-root step straight. As in the previous tutorial, begin with a valid distribution and find its mean. Then complete the deviation, squared-deviation, and weighted-squared-deviation columns. Add the final column to obtain variance, and take its square root to obtain standard deviation.

Procedure: For each possible value \(x\), calculate \(x-\mu_X\), then \((x-\mu_X)^2\), and then \((x-\mu_X)^2P(X=x)\). Add the final column to find \(\sigma_X^2\). Finally, calculate \(\sigma_X=\sqrt{\sigma_X^2}\).

Keep enough digits in intermediate calculations to avoid introducing avoidable rounding error. If the table entries produce an exact variance, take its square root and round the standard deviation at the end. If the variance is a decimal, enter the unrounded or full-precision value into the square root when using a calculator.

The columns also help catch errors. A negative deviation is possible, but its square and its weighted contribution cannot be negative. The final weighted contributions must add to the variance, not directly to the standard deviation. Only after finding that total should you take the square root.

1
Find the mean.
Calculate \(\mu_X=\sum xP(X=x)\), using every possible value and its probability.
2
Complete the variance columns.
For each row, calculate the deviation, square it, and multiply by the row’s probability.
3
Add and take the square root.
The sum of the weighted squared deviations is \(\sigma_X^2\); its nonnegative square root is \(\sigma_X\).
4
Report with units.
Variance has squared units; standard deviation has the same units as \(X\).

Worked Example: Reusable Containers Returned

Worked Example: Reusable Containers Returned

A campus dining program uses a model for the number of reusable containers returned by a randomly selected student during one lunch period. Let \(X\) be that number. Find the standard deviation.

Containers returned \(x\)\(P(X=x)\)
00.25
10.50
20.25

Check the distribution and find the mean. Each probability is between 0 and 1, and \(0.25+0.50+0.25=1.00\). The probability-weighted mean is:

$$ \begin{aligned} \mu_X &=0(0.25)+1(0.50)+2(0.25)\\ &=0+0.50+0.50\\ &=1.00\text{ container per student}. \end{aligned} $$

Complete the columns and find the variance. Use the mean of 1.00 for each deviation.

\(x\)\(P(X=x)\)\(x-\mu_X\)\((x-\mu_X)^2\)\((x-\mu_X)^2P(X=x)\)
00.25\(-1.00\)1.000.250
10.500.000.000.000
20.251.001.000.250

The weighted squared deviations sum to the variance:

$$ \sigma_X^2=0.250+0.000+0.250=0.500\text{ containers squared}. $$

Take the square root.

$$ \sigma_X=\sqrt{0.500}\approx 0.7071\text{ containers}. $$

State the result in context. According to this model, the standard deviation in the number of containers returned per student during one lunch period is about \(0.71\) container. The standard deviation is in containers, while the variance is in containers squared.

Worked Example: Missed Free Throws

Worked Example: Missed Free Throws

A coach models \(Y\), the number of free throws missed by a randomly selected player in a practice set of two attempts. The model assigns probability 0.50 to missing none, 0.30 to missing one, and 0.20 to missing both. Find the standard deviation.

Check the distribution and calculate its mean. The probabilities are between 0 and 1, and \(0.50+0.30+0.20=1.00\). Thus, the distribution is valid. Its mean is:

$$ \begin{aligned} \mu_Y &=0(0.50)+1(0.30)+2(0.20)\\ &=0+0.30+0.40\\ &=0.70\text{ missed free throws per set}. \end{aligned} $$

Use the mean to fill each calculation column.

\(y\)\(P(Y=y)\)\(y-\mu_Y\)\((y-\mu_Y)^2\)\((y-\mu_Y)^2P(Y=y)\)
00.50\(-0.70\)0.490.245
10.300.300.090.027
20.201.301.690.338

For example, the last row contributes \((2-0.70)^2(0.20)=1.30^2(0.20)=1.69(0.20)=0.338\). Add the contributions to get the variance, then take its square root:

$$ \begin{aligned} \sigma_Y^2&=0.245+0.027+0.338=0.610\text{ missed free throws squared},\\ \sigma_Y&=\sqrt{0.610}\approx 0.7810\text{ missed free throws}. \end{aligned} $$

State the result in context. Under the coach’s model, the standard deviation in the number of missed free throws per two-attempt set is about \(0.78\) missed free throw. This value is the square root of the variance, not the probability of missing a free throw.

Worked Example: Service Calls During a Shift

Worked Example: Service Calls During a Shift

A small repair shop models \(X\), the number of urgent service calls received during a randomly selected shift. Find \(\sigma_X\) and present a complete solution.

Urgent calls \(x\)\(P(X=x)\)
00.20
10.30
20.30
40.20

State. \(X\) is the number of urgent calls received during one shift. The requested quantity, \(\sigma_X\), is the standard deviation of this discrete random variable.

Plan. Check that the distribution is valid. Find \(\mu_X\), use it in the deviation and weighted-squared-deviation columns, add the last column to find \(\sigma_X^2\), and take the square root. The variance will be in calls squared per shift, and the standard deviation will be in calls per shift.

Do: check the distribution and calculate the mean. Every probability is between 0 and 1, and \(0.20+0.30+0.30+0.20=1.00\). The mean is:

$$ \begin{aligned} \mu_X &=0(0.20)+1(0.30)+2(0.30)+4(0.20)\\ &=0+0.30+0.60+0.80\\ &=1.70\text{ calls per shift}. \end{aligned} $$

Do: calculate every row’s contribution. The possible values are 0, 1, 2, and 4; include all four, even though the distribution has no row for 3.

\(x\)\(P(X=x)\)\(x-\mu_X\)\((x-\mu_X)^2\)\((x-\mu_X)^2P(X=x)\)
00.20\(-1.70\)2.890.578
10.30\(-0.70\)0.490.147
20.300.300.090.027
40.202.305.291.058

Add the weighted squared deviations, then take the nonnegative square root:

$$ \begin{aligned} \sigma_X^2 &=0.578+0.147+0.027+1.058\\ &=1.810\text{ calls squared per shift},\\ \sigma_X &=\sqrt{1.810}\\ &\approx 1.3454\text{ calls per shift}. \end{aligned} $$

Conclude. According to the shop’s model, the standard deviation in the number of urgent calls per shift is about \(1.35\) calls. It is the square root of the variance \(1.81\) calls squared, so it is expressed in the original units of calls per shift.

Common Mistakes and AP Exam Tips

  • Stopping at the variance. Adding the weighted squared deviations gives \(\sigma_X^2\), not \(\sigma_X\). A response that asks for standard deviation needs one more step: take the nonnegative square root.
  • Taking the square root of each row’s contribution. First add all values in the \((x-\mu_X)^2P(X=x)\) column. Then take the square root of the total. In general, the square root of a sum is not the sum of the square roots.
  • Using the wrong formula for a row. The weighted contribution is \((x-\mu_X)^2P(X=x)\): square the deviation, then multiply by the probability. Do not square the probability, and do not leave the deviation unsquared.
  • Reporting squared units for standard deviation. The variance has squared units; standard deviation has the same units as \(X\). Include the appropriate units when reporting each quantity.
  • Rounding too early. Carry enough digits through the table and take the square root of the unrounded variance when possible. Round the final standard deviation as appropriate to the context.
  • Confusing standard deviation with the range. The range is the difference between the largest and smallest possible values. Standard deviation comes from all values, their probabilities, and their distances from the mean; it is not found by subtracting endpoints.

For a full-credit response, show or clearly communicate the variance calculation, identify the square-root step, and report \(\sigma_X\) with the original units. If the question asks for an interpretation, identify what \(X\) measures and state the standard deviation in that context without calling it a guaranteed distance or a possible outcome.

Key Takeaway

Standard deviation is built directly from variance: calculate the weighted squared deviations in organized columns, add them to obtain \(\sigma_X^2\), and take the nonnegative square root. The square root changes the units from squared units back to the original units of the random variable.

Key takeaway: Find \(\sigma_X^2=\sum (x-\mu_X)^2P(X=x)\), then calculate \(\sigma_X=\sqrt{\sigma_X^2}\). Keep the variance and standard deviation distinct: variance has squared units, while standard deviation has the same units as \(X\).

Check Your Understanding

Use the distribution below, where \(X\) is the number of seedlings in a tray that sprout within a stated period.

Seedlings sprouted \(x\)\(P(X=x)\)
00.20
10.50
20.30
  1. Check that the table is a valid probability distribution.
  2. Calculate \(\mu_X\).
  3. Make a column table for \(x-\mu_X\), \((x-\mu_X)^2\), and \((x-\mu_X)^2P(X=x)\).
  4. Find \(\sigma_X^2\), then take its nonnegative square root to find \(\sigma_X\).
  5. State the units for the variance and standard deviation, and explain why they differ.

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