$P(S^2 > 3579)$?

Question

Suppose

$$\bar{X} = \frac{\sum_{i=1}^n}{n}$$

and

$$S^2 = \frac{1}{n-1} \sum_{i=1}^n (X_i - \bar{X})^2$$

Let $X_1, ... X_{20}$ a sample of size 20 with $N(300, 2500)$

1. The probability that the sample mean is less than $281.61$ is equal to $5$%.
2. The probability that the sample variance is greater than $3579$ is equal to $90$%.

For me, I think only 2 might be true. Intuitively, am I wrong?

I have tried $P(\bar{X} < 281.61) = 1 - \phi(\frac{300-281.61}{\sqrt{2500}}) = 1 - 0.64058 = 0.35942$. So the first choice can't be true. I have no clue where to start with $S^2$.

Answer 1

You can use the following statement (it is not difficult to prove it)

$$\frac{(n-1)S^2}{\sigma^2}\sim \chi_{(n-1)}^2$$

But the correct statement is the first.

$$\mathbb{P}[\overline{X}_{20}<281.61]=\mathbb{P}\Big[Z<\frac{281.61-300}{50}\sqrt{20}\Big]=\Phi(-1.64485)=0.05=5\%$$

The second statement is

$$\mathbb{P}[S^2>3579]=\mathbb{P}\Bigg[\chi_{(19)}^2>\frac{3579\times 19}{2500}\Bigg]=0.1=10\%$$

The first result has been derived using Z-table, the second using $\chi^2$ table