Worked Problems on Normal-Sample Variance and Studentization

From a Normal Sample to the Right Probability

Problems about a normal sample often ask for more than a distribution name. A probability involving $S^2$ requires the right chi-square degrees of freedom and scaling; a probability involving $\bar{X}$ may require standardization by either a known population standard deviation or the sample standard deviation. If both statistics appear, independence can sometimes reduce a joint probability to a product.

The key is to translate the event before evaluating it. This tutorial works through that translation for sample variances and studentized sample means. Every result below assumes a random sample of independent, identically distributed normal observations; without normality, the exact chi-square and Student t laws used here generally do not apply.

Definition: Let $X_1,\ldots,X_n$ be independent observations from $N(\mu,\sigma^2)$, where $n\geq 2$ and $\sigma^2>0$. The sample mean and unbiased sample variance are $$ \bar{X}=\frac{1}{n}\sum_{i=1}^{n}X_i, \qquad S^2=\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar{X})^2. $$ Under this normal-sample model, $(n-1)S^2/\sigma^2$ has a chi-square distribution with $n-1$ degrees of freedom. Also, $\bar{X}$ and $S^2$ are independent, and $\bar{X}\sim N(\mu,\sigma^2/n)$.

The Two Standardizations

When the population variance is known, standardize the sample mean with its standard error $\sigma/\sqrt{n}$. When the variance is unknown and the question uses the sample standard deviation $S$, standardize with $S/\sqrt{n}$; the resulting statistic has a Student t distribution with $n-1$ degrees of freedom.

Z=\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\sim N(0,1), \qquad T=\frac{\bar{X}-\mu}{S/\sqrt{n}}\sim t_{n-1}.

The chi-square statistic uses the sample variance itself:

Q=\frac{(n-1)S^2}{\sigma^2}\sim\chi^2_{n-1}.

These are exact distributions under the stated normal model, not approximations. In particular, the denominator of $S^2$ is $n-1$, so the degrees of freedom in the chi-square variable are also $n-1$. To convert a bound on $S^2$, multiply that bound by $(n-1)/\sigma^2$; do not change the scaling according to whether the question asks for an upper or lower tail.

Derivation: The normal-sample result gives $\bar{X}\sim N(\mu,\sigma^2/n)$, so $Z=(\bar{X}-\mu)/(\sigma/\sqrt{n})\sim N(0,1)$. It also gives $Q=(n-1)S^2/\sigma^2\sim\chi^2_{n-1}$, independently of $Z$. Therefore $$ \frac{Z}{\sqrt{Q/(n-1)}} = \frac{(\bar{X}-\mu)/(\sigma/\sqrt{n})} {\sqrt{S^2/\sigma^2}} = \frac{\bar{X}-\mu}{S/\sqrt{n}} =T. $$ A standard Student t variable is the ratio of an independent standard normal variable to the square root of an independent chi-square variable divided by its degrees of freedom. This gives $T\sim t_{n-1}$.

A Reliable Translation Process

For an integrated problem, first identify the model and the random quantities in the event. Then standardize each quantity separately, keeping track of whether the variance is known. If the event contains both $\bar{X}$ and $S^2$, use their independence only after checking that the sample is independent and normal.

Record the model and sample size.
Write $X_i$ independently distributed as $N(\mu,\sigma^2)$, and note $n$. This determines whether the exact sampling results apply.

Choose the statistic.
Use $(n-1)S^2/\sigma^2$ for a sample-variance event, $Z$ for a mean event with known $\sigma$, and $T$ for a studentized mean event.

Transform the event before looking up a probability.
Scale variance bounds by $(n-1)/\sigma^2$; standardize mean bounds by the appropriate standard error.

Evaluate and check.
Use the correct degrees of freedom and tail. For a joint event, multiply probabilities only when independence has been established and the event separates into one condition on each statistic.

Worked Problems

Worked Example: A Sample-Variance Threshold

A hypothetical sample consists of $n=5$ independent observations from $N(\mu,4)$, so the population variance is $\sigma^2=4$. Find $P(S^2<6)$. Give the probability to four decimal places.

Method. Use the exact chi-square law for the sample variance. There are $n-1=4$ degrees of freedom, and the event transforms as follows:

S^2<6 \quad\Longleftrightarrow\quad \frac{4S^2}{4}<\frac{4(6)}{4} \quad\Longleftrightarrow\quad \frac{(n-1)S^2}{\sigma^2}<6.

Thus the required probability is $P(\chi^2_4<6)$. For four degrees of freedom, the chi-square cumulative distribution function at $x\geq0$ is $1-e^{-x/2}(1+x/2)$. Substituting $x=6$:

P(S^2<6) =1-e^{-3}(1+3) =1-4e^{-3} \approx 0.8009.

Check. The threshold 6 is positive, and the resulting probability is between 0 and 1. The common scaling error is to use $6( n-1)/\sigma^2=4.5$. That is incorrect: $(n-1)/\sigma^2=4/4=1$, so the chi-square threshold is 6, not 4.5. The answer, rounded to four decimal places, is 0.8009.

Worked Example: A Joint Mean and Variance Event

A hypothetical sample has $n=5$ independent observations from $N(\mu,9)$. Find the probability that both $\bar{X}>\mu+1.5$ and $S^2>9$. Give the probability to four decimal places.

Method. Standardize each event. Since $\sigma=3$, the sample mean has standard error $3/\sqrt{5}$, and

P(\bar{X}>\mu+1.5) =P\left(Z>\frac{1.5}{3/\sqrt{5}}\right) =P\left(Z>\frac{\sqrt{5}}{2}\right) \approx 1-\Phi(1.1180) \approx 0.1318.

For the variance event, the chi-square degrees of freedom are $4$, and the threshold is

\frac{(n-1)(9)}{\sigma^2} =\frac{4(9)}{9} =4.

Therefore $P(S^2>9)=P(\chi^2_4>4)$. The chi-square survival probability with four degrees of freedom is $e^{-x/2}(1+x/2)$, so

P(S^2>9) =e^{-2}(1+2) =3e^{-2} \approx 0.4060.

Under the independent normal-sample model, $\bar{X}$ and $S^2$ are independent. The two event conditions concern these separate statistics, so their probabilities multiply:

P(\bar{X}>\mu+1.5,\ S^2>9) \approx 0.1318(0.4060) \approx 0.0535.

Check. The variance tail factor is $3e^{-2}$, not $5e^{-2}$. In particular, the scaled threshold is 4, and $P(\chi^2_4>4)=e^{-2}(1+2)$. The final joint probability, rounded to four decimal places, is 0.0535.

Worked Example: A Studentized Sample Mean

A hypothetical sample of $n=5$ independent observations is drawn from $N(\mu,\sigma^2)$, with $\sigma^2$ unknown. Find the probability that $\bar{X}-\mu>S/\sqrt{5}$. Give the probability to four decimal places.

Method. Divide the inequality by the positive quantity $S/\sqrt{5}$. The event becomes $T>1$, where $T=(\bar{X}-\mu)/(S/\sqrt{5})\sim t_4$. To evaluate this tail, the $t_4$ density is

f_{t_4}(t)=\frac{3}{8}\left(1+\frac{t^2}{4}\right)^{-5/2}.

Integrate from 0 to 1 and use symmetry about zero. With $y=t/2$, an antiderivative of $(1+y^2)^{-5/2}$ is $y/\sqrt{1+y^2}-y^3/[3(1+y^2)^{3/2}]$. Thus

P(T>1) =\frac{1}{2}-\frac{3}{4} \left[ \frac{y}{\sqrt{1+y^2}}- \frac{y^3}{3(1+y^2)^{3/2}} \right]_{0}^{1/2} =\frac{1}{2}-\frac{7}{10\sqrt{5}} \approx 0.1870.

Check. This is a one-sided upper-tail probability, so it is less than $0.5$, as symmetry requires. The answer is 0.1870, rounded to four decimal places. The unknown $\mu$ and $\sigma$ do not need numerical values because they cancel in the studentized statistic.

Reading the Result and Avoiding Traps

The transformation is often the most important part of an exam calculation. A chi-square probability is not obtained by comparing $S^2$ directly with a chi-square value: $S^2$ has units of squared observations, while $(n-1)S^2/\sigma^2$ is dimensionless. Likewise, a mean difference must be divided by a standard error, not by the population standard deviation alone.

Question involves	Statistic	Distribution and degrees of freedom
Sample variance $S^2$, with normal sampling	$(n-1)S^2/\sigma^2$	$\chi^2_{n-1}$
Sample mean, known population standard deviation	$(\bar{X}-\mu)/(\sigma/\sqrt{n})$	Standard normal
Sample mean studentized by $S$	$(\bar{X}-\mu)/(S/\sqrt{n})$	$t_{n-1}$

For a joint probability, independence is a special property of a normal sample: $\bar{X}$ and $S^2$ are independent under the independent, identically distributed normal model. It is not a general rule that a sample mean and sample variance from any population are independent. If the observations are not normal, the exact chi-square and t calculations shown here are not justified merely by having a large sample or by knowing the mean and variance.

Using $n$ degrees of freedom. The sample variance uses $n-1$ degrees of freedom, so its chi-square variable and the studentized mean both have $n-1$ degrees of freedom.
Scaling the variance threshold incorrectly. Transform $S^2<c$ or $S^2>c$ by multiplying both sides by $(n-1)/\sigma^2$. For example, if $n=5$ and $\sigma^2=4$, the multiplier is $4/4=1$.
Using the wrong standard error. A known-$\sigma$ mean calculation uses $\sigma/\sqrt{n}$. A studentized calculation uses $S/\sqrt{n}$, and its distribution is t rather than standard normal.
Multiplying joint probabilities without a reason. A product is appropriate for the separated mean and variance conditions here because of independence under normal sampling. It is not a shortcut that applies to arbitrary dependent events.

Key takeaway: State the normal-sample assumptions, identify the statistic, transform the event with the correct scale and degrees of freedom, and only then evaluate the tail. This sequence keeps chi-square variance calculations and Student t calculations aligned with the SOA VEE Mathematical Statistics syllabus.

Check Your Understanding

For each problem, write the distribution and degrees of freedom before evaluating the probability. Assume independent normal sampling wherever stated.

A sample of $n=7$ is drawn from $N(\mu,25)$. Find $P(S^2>50)$. (Answer: $P(\chi^2_6>12)=25e^{-6}\approx 0.0620$)
A sample of $n=5$ is drawn from $N(\mu,4)$. Find $P(S^2<6)$. (Answer: $P(\chi^2_4<6)=1-4e^{-3}\approx 0.8009$)
A sample of $n=9$ is drawn from $N(\mu,\sigma^2)$, where $\sigma$ is known. Find $P(\bar{X}-\mu>2\sigma/3)$. (Answer: $1-\Phi(2)\approx 0.0228$)
A sample of $n=6$ is drawn from a normal population with unknown mean and variance. Find $P((\bar{X}-\mu)/(S/\sqrt{6})>0)$. (Answer: $P(t_5>0)=0.5000$)
A sample of $n=5$ is drawn from $N(\mu,9)$. Find $P(\bar{X}>\mu+3/\sqrt{5}\text{ and }S^2<9)$. (Answer: $[1-\Phi(1)](1-3e^{-2})\approx 0.0942$)

Question involves	Statistic	Distribution and degrees of freedom
Sample variance \(S^2\), with normal sampling	\((n-1)S^2/\sigma^2\)	\(\chi^2_{n-1}\)
Sample mean, known population standard deviation	\((\bar{X}-\mu)/(\sigma/\sqrt{n})\)	Standard normal
Sample mean studentized by \(S\)	\((\bar{X}-\mu)/(S/\sqrt{n})\)	\(t_{n-1}\)

What You'll Learn