Mixed Probability Problems With T, Chi-Square, and F Distributions

One Problem-Solving Framework

A probability problem involving a t, chi-square, or F distribution has two distinct tasks. First, identify the random variable’s distribution and degrees of freedom from its definition. Then translate the event or requested quantile into a cumulative probability, a tail area, or a table lookup. Reversing these tasks can lead to using a convenient table entry for the wrong distribution.

The distributions are continuous, so a single exact value has probability zero. Probabilities concern intervals or tails, such as $P(T>c)$, rather than the probability that a variable equals precisely $c$. Tables often report selected upper-tail areas; software commonly provides cumulative probabilities and quantiles directly.

Definition: A standard t variable with $\nu$ degrees of freedom can be written $T=Z/\sqrt{U/\nu}$, where $Z\sim N(0,1)$, $U\sim\chi^2_\nu$, and $Z$ and $U$ are independent. A chi-square variable $U\sim\chi^2_\nu$ is a sum of squares of $\nu$ independent standard normal variables. An F variable with numerator degrees of freedom $d_1$ and denominator degrees of freedom $d_2$ can be written $F=(U_1/d_1)/(U_2/d_2)$, where $U_1\sim\chi^2_{d_1}$, $U_2\sim\chi^2_{d_2}$, and $U_1,U_2$ are independent. All degrees of freedom are positive.

These definitions also give a first check on an answer. A t variable can be negative or positive; chi-square and F variables are nonnegative. An F model requires two degrees of freedom, with the order specified: $d_1$ for the numerator and $d_2$ for the denominator. Changing that order generally changes the distribution.

From an Event to a Probability

Let $F_X(x)=P(X\le x)$ denote the cumulative distribution function (CDF) of a continuous random variable $X$. The CDF is a left-tail probability. An upper-tail probability is $P(X>x)=1-F_X(x)$. For a continuous variable, the choice of a strict or non-strict inequality at one endpoint does not change the probability.

Theorem: For a continuous random variable $X$ and $a<b$, $$ P(a<X\le b)=F_X(b)-F_X(a). $$ Consequently, $P(X>c)=1-F_X(c)$. For two nonoverlapping tails, $P(X<a\text{ or }X>b)=F_X(a)+1-F_X(b)$.

The interval formula follows by splitting the event $X\le b$ into two disjoint parts: $X\le a$ and $a<X\le b$. Therefore $F_X(b)=F_X(a)+P(a<X\le b)$; subtracting $F_X(a)$ gives the result. This reasoning applies to all three distributions. What changes from problem to problem is the CDF and its degrees of freedom.

Identify the variable.
Match its formula to a t, chi-square, or F definition. Check any stated independence and distribution assumptions.

Record the degrees of freedom.
Use $\nu$ for t or chi-square; for F, write both $d_1$ and $d_2$ in numerator–denominator order.

Rewrite the requested area.
Use a CDF difference for an interval, a complement for an upper tail, or the appropriate sum for disjoint tails.

Evaluate and interpret.
Use a table or software, account for rounding, and state what the calculated probability refers to.

Worked Problems With t and Chi-Square Distributions

Worked Example: A Two-Tail Probability for a t Variable

Suppose $Z\sim N(0,1)$ and $U\sim\chi^2_9$ are independent, and define $T=Z/\sqrt{U/9}$. Find $P(T<-2.262\text{ or }T>1.833)$.

Identify the model. The variable has the defining t form, with $\nu=9$ degrees of freedom. Independence is given, $Z$ is standard normal, and $U$ has nine chi-square degrees of freedom, so the conditions in the definition are met.

The t table gives $P(T>1.833)\approx0.050$ and $P(T>2.262)\approx0.025$ for $\nu=9$. The t distribution is symmetric about zero, so $P(T<-2.262)=P(T>2.262)$. The two events are disjoint, giving

P(T<-2.262\text{ or }T>1.833) \approx 0.025+0.050=0.075.

Thus the probability is approximately $0.075$, or $7.5\%$. The cutoffs are rounded table values, so software may give a slightly different result if those rounded numbers are used.

Worked Example: An Interval Probability and Quantile for Chi-Square

Let $X=Z_1^2+\cdots+Z_6^2$, where the $Z_i$ are independent standard normal variables. Find $P(1.237<X<12.592)$, and then find the 90th percentile of $X$.

Identify and check the model. This is a sum of squares of six independent standard normal variables, so $X\sim\chi^2_6$. The number of squared terms gives six degrees of freedom.

From a chi-square table for $df=6$, the lower-tail area at $1.237$ is approximately $0.025$, and the lower-tail area at $12.592$ is approximately $0.950$. Therefore,

P(1.237<X<12.592) =F_X(12.592)-F_X(1.237) \approx 0.950-0.025=0.925.

The probability that $X$ falls between these cutoffs is approximately $0.925$, or $92.5\%$. To find the 90th percentile, find $q$ such that $P(X\le q)=0.900$. The chi-square table or software gives $q\approx10.645$ for $df=6$. This is a lower-tail quantile: approximately $90\%$ of the distribution is at or below it.

For example, in R the corresponding calculations can be made with pchisq(12.592, 6) - pchisq(1.237, 6) and qchisq(0.90, 6). The first argument to qchisq is the requested lower-tail probability.

Worked Problems With F Distributions

Worked Example: An Upper-Tail Probability for F

Let $U_1\sim\chi^2_4$ and $U_2\sim\chi^2_{10}$ be independent, and define $F=(U_1/4)/(U_2/10)$. Find $P(F>2.605)$.

Identify and check the model. The variable is the ratio of two independent chi-square variables, each divided by its degrees of freedom. It therefore has an F distribution with numerator degrees of freedom $d_1=4$ and denominator degrees of freedom $d_2=10$. The numerator and denominator have not been interchanged.

An F table for $d_1=4$, $d_2=10$ gives an upper-tail area of approximately $0.10$ at $2.605$. Equivalently, the lower-tail CDF is approximately $0.90$, so

P(F>2.605)=1-P(F\le2.605)\approx1-0.90=0.10.

The probability is approximately $0.10$, or $10\%$. This is a table-based approximation because the cutoff is rounded. In R, pf(2.605, 4, 10, lower.tail = FALSE) directly requests the upper-tail probability.

Worked Example: Finding an F Quantile

For an $F$ distribution with $d_1=4$ and $d_2=10$, find the cutoff $c$ with $P(F>c)=0.05$.

Specify the model. The requested variable is $F\sim F_{4,10}$: the first degree of freedom belongs to the numerator and the second to the denominator. As in the definition, the underlying chi-square variables must be independent.

The requested upper-tail probability is $0.05$, so the equivalent lower-tail probability is $1-0.05=0.95$. An F table or software therefore gives the 95th lower-tail percentile for $F_{4,10}$, approximately $c=3.478$. In R, qf(0.95, 4, 10) returns this lower-tail quantile.

P(F>3.478)\approx 1-P(F\le3.478) =1-0.95=0.05.

Thus about $5\%$ of values from this F distribution exceed the cutoff. The answer is a quantile of the specified distribution, not a probability; its interpretation depends on both degrees of freedom.

Tables, Software, and Common Errors

A table lookup requires three pieces of information: the distribution, its degrees of freedom, and the tail convention used by the table. A chi-square table may label columns by upper-tail area, while software functions often default to the lower-tail CDF. An F table may likewise list critical values for specified upper-tail areas. Read the headings rather than assuming what a column means.

Software reduces interpolation and rounding, but it does not choose the correct model automatically. In R, pt(x, df) gives a lower-tail t probability, pchisq(x, df) gives a lower-tail chi-square probability, and pf(x, df1, df2) gives a lower-tail F probability by default. The corresponding quantile functions qt, qchisq, and qf take a lower-tail probability and degrees of freedom. An upper-tail request can be obtained either by taking one minus a lower-tail probability or by using the software’s documented upper-tail option.

For a quantile, be explicit about which tail is specified. If $P(X>c)=0.05$, then $P(X\le c)=0.95$ for a continuous distribution. Feeding $0.05$ to a lower-tail quantile function would instead return a cutoff with only $5\%$ below it.

Wrong reasoning: “The table gives $0.05$, so the probability below the cutoff is $0.05$.” Correction: First check whether that table entry is a lower-tail or upper-tail area.
Wrong reasoning: “For $F_{4,10}$, the order of the degrees of freedom does not matter.” Correction: The numerator and denominator degrees of freedom are distinct; swapping them generally changes the distribution and the probability.
Wrong reasoning: “A chi-square or F variable can be negative if the observed value is unusual.” Correction: Both are nonnegative by definition; a negative value signals a mistaken formula or calculation.

These checks are useful because an arithmetic step can be flawless while the final probability is wrong: a reversed tail or an incorrect degree-of-freedom order changes the question being answered. Write down the model, parameters, and tail before entering values into a table or software.

Key takeaway: Identify the distribution from its formula, carry its degrees of freedom carefully, and convert the requested event into a lower-tail CDF, a complement, or a CDF difference before calculating.

Check Your Understanding

For each question, identify the distribution and degrees of freedom before deciding how to calculate the requested quantity.

If $Z\sim N(0,1)$, $U\sim\chi^2_7$, and $Z$ and $U$ are independent, what is the distribution of $Z/\sqrt{U/7}$?
A variable is the sum of five squared independent standard normal variables. State its distribution and degrees of freedom.
For $X\sim\chi^2_8$, explain how to express $P(a<X\le b)$ using the CDF $F_X$.
For $F\sim F_{6,12}$, what does $P(F>c)=0.02$ imply about $P(F\le c)$?
Why is the F distribution for $d_1=6,d_2=12$ generally different from the one for $d_1=12,d_2=6$?