Find x in Standard Deviation: Formula and Examples

This calculator finds the standard deviation of a set of numbers you enter, showing how spread out the values are around their mean. Enter the data values as a comma-separated list to get the population standard deviation; the steps below also explain how to find an unknown value x when a standard deviation is given.

Formula

For a population of N values, the population standard deviation is σ = √[Σ(xᵢ − μ)² / N]. Here, xᵢ is each data value, μ is the population mean, N is the number of values, and σ is the population standard deviation. All values and the result use the same units.

For a sample, use s = √[Σ(xᵢ − x̄)² / (n − 1)], where x̄ is the sample mean and n is the number of observations. The calculator reports the population version, dividing by N. It does not use the sample correction.

When solving for an unknown value x from a specified standard deviation, start with the relevant formula and isolate x. The result may have two possible values because standard deviation measures squared distance from the mean: a value can lie above or below it by the same amount.

Worked Example

Using the default list 2, 4, 4, 4, 5, 5, 7, 9:

1. There are N = 8 values.
2. Their sum is 40, so the mean is μ = 40 / 8 = 5.
3. The squared deviations from 5 are 9, 1, 1, 1, 0, 0, 4, and 16. Their sum is 32.
4. The population variance is 32 / 8 = 4, and the standard deviation is √4 = 2.

The calculator therefore reports a population standard deviation of 2 units, 8 values, and a mean of 5 units.

What You’ll Need

• A list of numeric observations, entered with commas between values.
• If solving for x, the known mean, standard deviation, other data values, and whether the problem describes a population or a sample.
• A calculator for squaring values and taking a square root, or the calculator above for a complete list.

Step-by-Step

1. Identify what is unknown. “Find x in standard deviation” can mean calculating the standard deviation of a data set that contains x, or recovering x from a known standard deviation. Check the wording and note whether the formula is for a population or sample.
2. Write down the data and known quantities. Record every observed value, the stated mean if supplied, the standard deviation, and the number of observations. Keep units consistent. If the mean is not given, calculate it as the sum of the values divided by their count.
3. Choose the correct formula. For a population, use σ² = Σ(xᵢ − μ)² / N. For a sample, use s² = Σ(xᵢ − x̄)² / (n − 1). If x is one of the data values, its contribution to the sum is (x − mean)².
4. Substitute known values. If the mean and standard deviation are known, square the standard deviation to obtain the variance. Then write an equation for the squared deviations. For example, with known mean μ and one unknown x, its squared deviation is (x − μ)².
5. Isolate the unknown term. Subtract the known squared deviations from the total squared-deviation sum. For a population, that total is Nσ²; for a sample it is (n − 1)s². This leaves the squared deviation for x, provided the other values and the applicable divisor are known.
6. Take the square root and include both signs. If (x − μ)² = A, then x − μ = ±√A, so x = μ ± √A. Check both values against any conditions in the question, such as a restricted range or a requirement that x be greater than the mean.
7. Check the result in the original formula. Substitute each allowed candidate for x, recalculate the mean if it was not fixed, and verify that the standard deviation matches the stated value. Round only at the end.

For example, if the mean is 10 and an unknown value is known to be exactly 3 units from the mean, then (x − 10)² = 9. Taking square roots gives x − 10 = ±3, so x may be 7 or 13. This is a simple illustration; a data set with several unknowns or an unknown mean requires solving the full equation, not just this single-deviation step.

Common Mistakes

• Confusing standard deviation and variance: variance is in squared units; standard deviation is its square root and uses the original units.
• Using the wrong divisor: population calculations divide by N; sample calculations divide by n − 1. Follow the problem’s description.
• Dropping the negative root: taking a square root of a squared deviation usually gives two possibilities, not one.
• Assuming x is the only unknown: a standard deviation alone may not determine a unique x, especially if the mean or other data values are also unknown.
• Rounding too early: intermediate rounding changes the final result. Keep full calculator precision until the last step.
• Mixing units or entering a malformed list: convert all measurements to one unit and separate list values with commas.

Tips

• Draw a quick number line around the mean. Candidate values that give the same standard deviation are often equally distant on opposite sides.
• Use squared deviations, not signed deviations. Positive and negative differences would cancel if you simply added them.
• Keep a record of whether the question uses a sample or a full population; the divisor affects the answer.
• For a list, verify the count and mean before trusting the final standard deviation.
• Remember that standard deviation is sensitive to extreme values because deviations are squared.

Troubleshooting

The equation has no real solution: check the divisor and arithmetic. The amount left for an unknown squared deviation cannot be negative. A negative remainder often means the stated standard deviation or another supplied value is inconsistent, or a sample/population formula was mixed up.

You get two values for x: that is normal when only distance from the mean is known. Use additional information in the question to decide whether one or both candidates are valid.

Your recalculated result does not match: check whether you recomputed the mean after inserting x, included every observation, and used the correct number of values. Avoid rounding intermediate values.

The calculator output looks wrong: inspect the list for omitted values, accidental spaces used in place of separators, or values in different units. The calculator computes population standard deviation; for a sample, apply the sample formula separately.

Frequently Asked Questions

Can standard deviation alone tell me x?
No. You generally also need the mean and enough information about the other data values. A known deviation from a fixed mean often gives two possible x values.

Why do I get plus or minus when solving for x?
Because squaring removes the sign: values equally far below and above the mean have the same squared deviation.

Should I use n or n − 1?
Use N for a complete population. Use n − 1 for a sample when estimating spread from a sample, unless the problem specifies another convention.

Can standard deviation be negative?
No. It is the nonnegative square root of a variance. A negative result signals an error or that a different quantity was calculated.

What happens if all values are the same?
Every value equals the mean, so all deviations are zero and the standard deviation is zero.

Does the calculator find an unknown x automatically?
It calculates population standard deviation from a complete list. To solve for an omitted x, use the known quantities and algebraic steps above, then verify the candidate in the formula.

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list data | Data values | 2, 4, 4, 4, 5, 5, 7, 9
out sd | Population standard deviation | units | psd(data) | 2
out n | Number of values | - | count(data) | 0
out mean | Mean | units | mean(data) | 2