Proporties of linear-combinations of normally distributed variables.

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At my school, to pass an exam, you'll have to score at least 230 points. The results are normally distributed with $\mu=200$ and $\sigma=20$.

If I were to consider 10 students who attends the exam, where they achieve $x_1, x_2,...,x_{10}$ points.

a) Whats the probability that only one student gets the distinction mark?

I know that I can use the $x_i's$ as a sample from the described population. If the $x_1,x_2,...,x_{10}$ are normally distributed with $X_i\sim N(\mu=200,\sigma=20)$, right?

But all of the $x_i's$ share the same mean value and standarddeviation which implies that $\sum X_i\sim (10\cdot 200, 10\cdot 20^2)$ . What's the next step now? I tried to define $\bar{X}=\frac{1}n\sum X_i$ so $\bar{X}=\frac{1}{10} \sum X_i$

This leaves me with $\bar{X}\sim N(\mu,\frac{\sigma^{2}}{n})$ which equals $\bar{X}\sim N(200,40)$

This is where I'm stuck. If I want to find the probability of exactly one student getting the $≥230$ points, how should I continue?

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The probability of passing is an independent bernoulli variable with probability $$p_i = P(X_i \ge 230)$$ Thus the probability for exactly one out of $10$ is $$\binom{10}1 (1-p)^9 p^1 = 10 (1-p)^9p$$