Chi distribution and sample variance

Question

Suppose that the height (in cm) of randomly selected male is distributed according to normal distribution with parameters $\mu = 175$ and $\sigma = 5$. We pick a simple random sample of size $101$ from population of males. What is the value such that the probability, that the sample variance is larger than this value, is $0.05$?

Solution attempt:

For a normal population, the quantity $${{(n-1)s^2} \over{\sigma^2} }$$ is known to have a chi-squared distribution with $n-1$ degrees of freedom.

So then $$P[{s^2} > X] = P\left[ {\frac{{(101 - 1){s^2}}}{{{5^2}}} > X} \right] = P\left[ {\chi^2_{(100)}> X} \right] = 0.05$$

Using $R$ command: qchisq(p = 0.05, df = 100, lower.tail = FALSE) i found that $$X \approx 124.3421$$

Have i solved this problem correctly?

Answer 1

There is a mistake in the equation $$P[s^2>X]=P\left[\frac{(101-1)s^2}{5^2}>X\right]$$ You need to multiply both sides of the inequality by $\frac{101-1}{5^2}$, not just the left side.