Sampling Distribution of Variances Another Exmaple

Question

Below is a problem I did but my answer does not match the book. The problem is from the book "Probability and Statistics" which is one of Schaum's books. I am using the second edition. I am hoping somebody can tell me where I went wrong or that the book is wrong. I suspect the book is right.
Thanks
Bob
Problem:
A normal population has a variance of $15$. If samples of size $5$ are drawn from this population, what percentage are be expected to have variances more than $20$.
Answer
Let $p_{20}$ be the probability that the sample variance is under $20$. Let $p$ be the probability we seek. Now we have $p = 1 - p_{20}$. To find $p_{20}$ we go to the online calculator with $4$ degrees of freedom. The chi-square critical value is $4(\frac{20}{15}) = 5.3333333$. From the calculator we get: $p_{20} = 0.75$. Hence, $p = 0.25$.
Note: The calculator can be found at this URL:
      https://stattrek.com/online-calculator/chi-square.aspx
The book gets: $p = 0.17$.

Answer 1

If $X_1, X_2, \dots X_5$ is a random sample from $\mathsf{Norm}(\mu, \sigma=\sqrt{15}),$ then $Q = 4S^2/15 \sim \mathsf{Chisq}(4),$ where $S^2$ is the sample variance.

Thus, $$P(S^2 > 20) = P\left(\frac{4S^2}{15} > \frac{4(20)}{15} = 5.3333\right) = 0.2548.$$ Computation in R statistical software:

1 - pchisq(16/3, 4)
[1] 0.2547727

Note: The simulation below with a million such samples of size $n=5$ agrees with this result within the margin of simulation error. (Two or three place accuracy can be expected.)

set.seed(918)
v = replicate( 10^6,  var(rnorm(5, 0, sqrt(15))) )
mean(v > 20)
[1] 0.25556