Calculating Probabilities for a Normal Sample Variance

From Sample-Variance Bounds to Probabilities

A sample variance is measured in the same squared units as the underlying observations, but its sampling distribution is not usually described directly in those units. For an independent sample from a normal population, rescaling the sample variance produces a chi-square random variable. This gives a direct method for answering questions such as the probability that the sample variance exceeds a specified threshold, or falls between two variance values.

The calculation has two important parts: translate each variance bound to the chi-square scale, and use the correct degrees of freedom. Reversing a tail or substituting the population standard deviation for the population variance changes the result.

Definition: Let $X_1,\ldots,X_n$ be an independent random sample from $N(\mu,\sigma^2)$, with $\sigma^2>0$. Define the sample mean and unbiased sample variance by $$ \bar{X}=\frac{1}{n}\sum_{i=1}^n X_i, \qquad S^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar{X})^2. $$ Then $$ \frac{(n-1)S^2}{\sigma^2}\sim\chi^2_{n-1}. $$ The degrees of freedom are $n-1$, and the scale in the denominator is the population variance $\sigma^2$, not the population standard deviation $\sigma$.

The Basic Probability Conversion

Write $F_{\chi^2_\nu}(x)=P(W\leq x)$ when $W\sim\chi^2_\nu$. If $a$ is a proposed sample-variance cutoff, multiply it by $(n-1)/\sigma^2$ to obtain the corresponding cutoff for $W$. Since this multiplier is positive, the direction of each inequality is preserved.

W=\frac{(n-1)S^2}{\sigma^2},\qquad W\sim\chi^2_{n-1}

For example, a lower-tail probability becomes $P(S^2\leq a)=F_{\chi^2_{n-1}}\big((n-1)a/\sigma^2\big)$. An upper-tail probability is one minus that CDF value. For an interval, subtract the CDF at the transformed lower bound from the CDF at the transformed upper bound.

\begin{aligned} P(S^2\leq a) &=F_{\chi^2_{n-1}}\left(\frac{(n-1)a}{\sigma^2}\right),\\ P(S^2>a) &=1-F_{\chi^2_{n-1}}\left(\frac{(n-1)a}{\sigma^2}\right),\\ P(a<S^2<b) &=F_{\chi^2_{n-1}}\left(\frac{(n-1)b}{\sigma^2}\right) -F_{\chi^2_{n-1}}\left(\frac{(n-1)a}{\sigma^2}\right). \end{aligned}

These expressions use the chi-square CDF, which can be obtained from a suitable table or statistical software. For even degrees of freedom, some values can also be computed using a finite sum. If $W\sim\chi^2_{2m}$ and $x\geq0$, then $F_{\chi^2_{2m}}(x)=1-e^{-x/2}\sum_{j=0}^{m-1}(x/2)^j/j!$. The examples use this form to show the numerical steps.

Why the conversion works: For every outcome, $W=(n-1)S^2/\sigma^2$. If $c=(n-1)/\sigma^2>0$, then $S^2\leq a$ if and only if $cS^2\leq ca$. Therefore the event about $S^2$ is exactly the corresponding event about $W$. Applying the chi-square distribution function to the converted cutoff gives the probability. For an interval, the probability is the upper cumulative probability minus the lower cumulative probability.

Worked Probability Calculations

Worked Example: Probability of a Sample Variance in an Interval

A hypothetical laboratory takes an independent sample of $n=5$ measurements from a normal population with known population variance $\sigma^2=9$ measurement-units squared. Find the probability that the unbiased sample variance is between $4.5$ and $18$ measurement-units squared.

Method. Standardize both variance bounds using $W=(n-1)S^2/\sigma^2$. Here the chi-square degrees of freedom are $5-1=4$.

The transformed lower and upper bounds are $\frac{4(4.5)}{9}=2$ and $\frac{4(18)}{9}=8$. Thus $P(4.5<S^2<18)=P(2<W<8)$, where $W\sim\chi^2_4$.

For four degrees of freedom, $F_{\chi^2_4}(x)=1-e^{-x/2}(1+x/2)$. At the upper bound, $F_{\chi^2_4}(8)=1-e^{-4}(1+4)=1-5(0.0183156)=0.908422$. At the lower bound, $F_{\chi^2_4}(2)=1-e^{-1}(1+1)=1-2(0.3678794)=0.264241$.

Subtract the cumulative probabilities: $P(2<W<8)=0.908422-0.264241=0.644181$. The probability is 0.6442, or about 64.42%. In the original units, this is the probability that repeated samples of size five produce a variance between $4.5$ and $18$ measurement-units squared under the stated normal model.

Check and trap avoided. Both converted cutoffs are increasing with the original variance bound, and the upper CDF value is larger than the lower one. Subtracting in the reverse order would give an impossible negative probability.

Worked Example: An Upper-Tail Sample-Variance Probability

A hypothetical portfolio analyst models a claim-related measurement as normally distributed with population variance $\sigma^2=16$ units squared. An independent sample of $n=7$ observations is collected. Find the probability that its unbiased sample variance exceeds $32$ units squared.

Method. Convert the event to an upper-tail chi-square probability. The degrees of freedom are $7-1=6$, and the standardized cutoff is $\frac{6(32)}{16}=12$. Hence the required probability is $P(W>12)$ for $W\sim\chi^2_6$.

For six degrees of freedom, $F_{\chi^2_6}(x)=1-e^{-x/2}\left(1+x/2+(x/2)^2/2\right)$. At $x=12$, this gives $F_{\chi^2_6}(12)=1-e^{-6}(1+6+18)=1-25(0.00247875)=0.938031$.

The requested probability is the right tail: $P(W>12)=1-0.938031=0.061969$, or 0.0620 to four decimal places. Thus, under the model, about 6.20% of samples of this size would have sample variance greater than $32$ units squared.

Check and trap avoided. The event asks for a variance greater than the cutoff, so its probability is the right tail, not the CDF value $0.9380$. Also, the denominator in the scaling is the population variance $16$, not its square root, $4$.

Worked Example: Translating an Interval with More Degrees of Freedom

A hypothetical quality-control team observes $n=9$ independent measurements from a normal process with population variance $\sigma^2=25$ units squared. Find the probability that the unbiased sample variance is between $12.5$ and $37.5$ units squared.

Method. The transformed variable has $9-1=8$ degrees of freedom. The lower chi-square bound is $\frac{8(12.5)}{25}=4$, and the upper bound is $\frac{8(37.5)}{25}=12$. The desired probability is $P(4<W<12)$, where $W\sim\chi^2_8$.

For eight degrees of freedom, $F_{\chi^2_8}(x)=1-e^{-x/2}\left(1+x/2+(x/2)^2/2+(x/2)^3/6\right)$. At $x=4$, $F_{\chi^2_8}(4)=1-e^{-2}(1+2+2+1.333333)=1-0.857123=0.142877$. At $x=12$, $F_{\chi^2_8}(12)=1-e^{-6}(1+6+18+36)=1-0.151204=0.848796$.

Subtract to get $P(4<W<12)=0.848796-0.142877=0.705919$, or 0.7059 to four decimal places. In context, this is the model probability that the sample variance falls in the stated interval of squared measurement units.

Check and trap avoided. The original bounds are in variance units, so they are multiplied by $8/25$, not divided by the population standard deviation. The transformed interval remains ordered: $4<12$.

Assumptions, Interpretation, and Common Errors

The chi-square conversion is exact when the observations are an independent random sample from one normal population with a common mean and variance, and $S^2$ is defined with denominator $n-1$. These conditions are part of the model, not optional details. If the population is not normal, the scaled sample variance does not in general have this exact chi-square distribution. A large sample alone does not make the exact identity true.

The calculation returns a probability about the sampling behavior of $S^2$, not the probability that the fixed population variance changes or that a particular observed sample variance is correct. For example, a result of $0.0640$ means that about 6.4% of repeated samples under the model would produce a sample variance in the specified event.

Keep the units visible while standardizing. If observations are measured in dollars, both $S^2$ and $\sigma^2$ are in dollars squared, so their ratio is dimensionless, as a chi-square variable must be. If a problem gives the population standard deviation $\sigma$, square it before using the formula.

Confirm the model and definition.
Check normality, independence, common variance, and the $n-1$ denominator in $S^2$.

Set the degrees of freedom.
Use $\nu=n-1$, not $n$. The loss of one degree of freedom comes from estimating the population mean with the sample mean.

Convert every variance cutoff.
Multiply each bound $a$ by $(n-1)/\sigma^2$. Keep the bounds in increasing order.

Match the event to the tail or interval.
Use a CDF for a lower tail, one minus the CDF for an upper tail, or upper CDF minus lower CDF for an interval.

Interpret in the original context.
State that the probability concerns repeated samples and refers to the original sample-variance bounds.

A frequent error is to replace $(n-1)S^2/\sigma^2$ with $nS^2/\sigma^2$. That replacement is incorrect for the unbiased sample variance defined here. Another is to report a lower-tail CDF when the event is an upper tail. A quick check is to ask whether the event is “below,” “above,” or “between” the original cutoffs before consulting the table or software.

Key takeaway: Under the exact normal-sample conditions, translate a sample-variance cutoff $a$ to $(n-1)a/\sigma^2$, use $n-1$ chi-square degrees of freedom, and then evaluate the correctly directed tail or interval. The result remains a probability about $S^2$ in its original variance units.

Check Your Understanding

For each problem, use the unbiased sample variance and round probabilities to four decimal places.

An independent sample of size $3$ comes from a normal population with variance $\sigma^2$. Find the probability that $S^2>2\sigma^2$. Interpret the probability as a repeated-sampling statement. (Answer: 0.1353)
A normal population has variance $4$ units squared. For an independent sample of size $5$, find $P(S^2\leq 6)$. (Answer: 0.8009)
A normal population has variance $9$ units squared. For an independent sample of size $7$, find the probability that $S^2$ lies between $4.5$ and $13.5$ units squared. (Answer: 0.6353)
A normal population has variance $25$ units squared. An independent sample of size $5$ is taken. Find the probability that its sample variance exceeds $50$ units squared. (Answer: 0.0916)
A normal population has variance $9$ units squared, and an independent sample of size $5$ is taken. Find the probability that $S^2<3$ or $S^2>9$ units squared. (Answer: 0.5503)

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