If the number of persons per car is a random variable $H$ with mean $4$ and variance $2$, what is $P(\bar H\geq 5)$?

Question

The question is as follows:

According to a transportation safety board, the number of persons per car passing a certain intersection between 8:00 and 9:00am, is a random variable $H$ with mean $4$ and variance $2$.

For a random sample of $30$ cars at this intersection during this time period, what is the probability that the average number of persons per car is at least $5$?

I've figured out that I'll be using the Central Limit Theorem to solve this, and thus far have found that: $$P(H > 5) = 1 - P( H \leq 5).$$

But I can't seem to incorporate the sample of $30$ cars in my equation, this would only be true for a sample $n$ of $1$ car.

How can I work from here?

Answer 1

I believe we can proceed as follows.

Presumably, we have that $H_1,\dotsc,H_{30}$ are iid with $E[H_i] = 4$, and $V(H_i) = 2$.

Notice that the average is $$\bar H = \frac{H_1+\dotsb+H_{30}}{30}$$ and hence the question asks to find $P(\bar H\geq 5)$. So, we compute $$E[\bar H] = E\left[\frac{H_1+\dotsb+H_{30}}{30}\right] = \frac{1}{30}(E[H_1]+\dotsb+E[H_{30}]) = \frac{1}{30}\cdot 30 E[H_1] = 4$$ and similarly, by independence, $$V(\bar H) = \frac{1}{30^2}(V(H_1)+\dotsb+V(H_{30})) = \frac{1}{30^2}\cdot30\cdot 2 = \frac{2}{30}.$$

Now, we proceed as usual $$P(\bar H\geq 5) \approx P\left(Z\geq \frac{5-4}{\sqrt{2/30}}\right) = 1- P\left(Z< \frac{5-4}{\sqrt{2/30}}\right) = 1-\Phi\left(\frac{5-4}{\sqrt{2/30}}\right)$$ where $Z$ is a standard normal, and $\Phi$ is the usual cdf function of $Z$.

Answer 2

The idea of the Central Limit Theorem is that the if $X_1,\ldots,X_n$ (with $n$ "large enough") are random variables with the same distribution, and $E(X)=\mu$ and $Var(X)=\sigma^2$, then the new random variable $Z=\dfrac{\overline{X}-\mu}{\sigma/\sqrt{n}}$ has normal distribution with mean 0 and variance 1.

In your case, you have $H_1,\ldots,H_{30}$ with the same distribution, $\mu=4$, $\sigma=\sqrt{2}$ and you do

$$P(\overline{H}\geq 5)=P\left(\dfrac{\overline{H}-\mu}{\sigma/\sqrt{n}}\geq \dfrac{5-4}{\sqrt{2}/\sqrt{30}}\right)=P(Z\geq 3.87)$$ and it is the last probability the one that you look either on a table of values for the normal distribution, or in some software that you might know.

Answer 3

Let $H_1$ be the number of people in the first car in the sample, $H_2$ the number of people in the second car, and so on up to $H_{30}$.

Let $X=\frac{1}{30}(H_1+H_2+\cdots+H_{30})$. Then random variable $X$ is the average number of people per car in the sample.

We want the probability that $X\ge 5$. Note that $X$ has mean $4$ and variance $2/30$.

Since $X$ is a sum of a not too tiny number of identically distributed random variables, it is not too unreasonable to assume that $X$ has a not far from normal distribution.

So we want to find the probability that a normally distributed random variable with mean $4$ and variance $\frac{2}{15}$ is $\ge 5$. I expect you have computed this sort of probability a number of times.

Remark: The normal approximation will give that the probability is close to $0$. The estimate should not be taken entirely seriously, since the approximation error of $X$ by a normal is undoubtedly greater than the small number we obtained.

Answer 4

Just to look at it a different way, you can treat the sum of the people in the $30$ cars as the random variable, which has mean $30 \times 4 = 120$ and variance $30 \times 2 = 60$. It therefore has standard deviation $\sqrt{60} \doteq 7.75$.

A mean of $5$ implies a total of $150$ or more people in the $30$ cars, which gives

$$ Z = \frac{150-120}{\sqrt{60}} = \sqrt{15} \doteq 3.87 $$

and you can now proceed as in the other answers.