Basic question about using the chi-square table We are struggling on this easy question : If X is a chi-square random variable with 6 degrees of freedom, find $P(X \le 6)$
We know the answer is 0.58 according to some online calculator. We need to find it.
On the table, we go to line with n = 6. However, 6 is somewhere in the large interval of 1.635 (when alpha = 0.95) and 12.592 (when alpha = 0.05)
How do we come up with 0.58?
Thanks!
 A: Here’s one way to answer your question analytically (no tables).  The value of $P(X \le 6)$ in your problem equals the cumulative distribution function of the $\chi^2$-distribution with $k=6$ degrees of freedom for the value $x=6$.  This is given by the expression
$$\frac{\gamma(\frac{k}{2},\,\frac{x}{2})}{\Gamma(\frac{k}{2})}$$
where $\gamma$ is the lower incomplete gamma function and $\Gamma$ is the ordinary gamma function.
The denominator is just $\Gamma(3)=2!$.  The numerator is given by 
$$\gamma (3,3) = \int_0^3 {t^2 } e^{ - t} dt.$$
Using integration by parts, we find that $\gamma (3,3) = 2 - 17e^{ - 3}$.  Therefore, we have $P(X \le 6)=1-\frac{17}{2}e^{-3}\approx 0.576809919$.
A: Most tables for the chi-square distribution are not designed to give you general probabilities; they are designed to give you critical values for specific tail probabilities corresponding to various significance levels.  For example,  refer to the following table:

To find $\Pr[X \le 6]$ using this table, you'd look up the sixth row, and try to find the column for which the entry in that table equals $6$.  In other words, the sixth row and fifth column of this table means $\Pr[X > 5.348] \approx 0.5$, and the sixth row and seventh column means $\Pr[X > 7.84] \approx 0.25$.  So in order to get $\Pr[X \le 6] = 1 - \Pr[X > 6]$, we would need a column somewhere in between $0.5$ and $0.25$, but it's not there in the table.
We can, however, use a crude linear interpolation:  If we know that $\Pr[X > 5.348] = 0.5$ and $\Pr[X > 7.84] = 0.25$, then we can estimate that $$\Pr[X > 6] \approx 0.5 (1-\lambda) + 0.25 \lambda,$$ where $$ \lambda = \frac{6 - 5.348}{7.84 - 5.348} \approx 0.261637.$$  This gives $\Pr[X > 6] \approx 0.434591$, which gives $\Pr[X \le 6] \approx 0.565409$.  It's not that bad an approximation; the actual answer calculated with a computer is $\Pr[X \le 6] = 0.57681\ldots$.
