Normal distribution - How to Calculate When There are Several Variables?

Question

I have this task:

At a gathering, the contestants are supposed to taste a new product. The time a random contestant uses to taste this product is normally distributed with $\mu = 9\text{ minutes}$ and $\sigma = 3\text{ minutes}$.

a) Five random contestants are sitting around table $1$. What is the probability that they on average use less than $8$ minutes?

b) What is the probability that exactly four of them use less than $8$ minutes?

I know how to do this for one person, but what do I need to take into consideration when doing this for a group of people?

Thank you!

Answer 1

You know that $$X_i\sim\mathrm{N}\left(9,3^2\right)$$ for $i=1,2,3,4,5$.

$$\overline{X}=\frac{X_1+X_2+X_3+X_4+X_5}5$$

Hence,

\begin{align} E\left[\overline{X}\right]&=E\left[\frac{X_1+X_2+X_3+X_4+X_5}5\right]\\ &=\frac{E\left[X_1\right]+E\left[X_2\right]+E\left[X_3\right]+E\left[X_4\right]+E\left[X_5\right]}{5}\\&=\mu=9\end{align}

and

\begin{align}\mathrm{Var}\left[\overline{X}\right]&=\mathrm{Var}\left[\frac{X_1+X_2+X_3+X_4+X_5}5\right]\\&=\frac{\mathrm{Var}\left[X_1\right]+\mathrm{Var}\left[X_2\right]+\mathrm{Var}\left[X_3\right]+\mathrm{Var}\left[X_4\right]+\mathrm{Var}\left[X_5\right]}{5^2}\\&=\frac{\sigma^2}5=\frac{3^2}{5}=\left(\frac{3}{\sqrt5}\right)^2\end{align}

Therefore, $$\overline{X}\sim\mathrm{N}\left(9,\left(\frac{3}{\sqrt5}\right)^2\right)$$

Your probability for (a) is $$P\left(\overline{X}<8\right)$$

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For (b), if you know the binomial distribution, the number of contestants that uses less than $8$ minutes, $Y$, is distributed by $$Y\sim\mathrm{B}\left(5,P\left({X_i}<8\right)\right)$$

Otherwise, these are the steps:

1. Choose one contestant to use more than or equal to $8$ minutes. There are $$\binom{5}{1}$$ ways to do this.
2. Find the probability of that contestant using more than or equal to $8$ minutes. $$P\left({X_a}\ge8\right)$$
3. Find the probability of the other four contestants using less than $8$ minutes. $$P\left({X_b}<8\right)P\left({X_c}<8\right)P\left({X_d}<8\right)P\left({X_e}<8\right)$$

Hence, the required probability is $$\binom{5}{1}P\left({X}\ge8\right)\left(P\left({X}<8\right)\right)^4$$