Expected value of random variable $X$ ~ $N(170, 25)$

Question

Here's a question:

Person's height in CMs is a random variable $X$ ~ $N(170, 25)$. Door's height is $180$ cm. What is the expected value of number of people that can enter the door until the first one who will need to bend ?

To be honest I can't think about a way to solve it. Can you help me with that? Thanks in advance.

Answer 1

HINT:

Let $k$ denote the probability that a person can walk through the door without bending:

$$k=P(X\leq180)$$

The probability that $n$ people will walk through the door until the first one bends is:

$$k^{n-1} \cdot (1-k)^1$$

The expected number of people that will walk through the door until the first one bends is:

$$\sum\limits_{n=1}^{\infty}k^{n-1} \cdot (1-k)^1 \cdot n$$

Answer 2

Based on the Normal distribution given , find the probability that a person's height exceeds $180$ cm, call this $P(X>180)$. The expected number of people that can enter the door will be the reciprocal of this, i.e. $1/P(X>180)$

Answer 3

Let $Y$ be the number of people who enter until the first one has to bend. Note that $Y$ is discrete. Take $k \in \mathbb{N}$. If $Y=k$ this means that $k$th person to go through the door bends while the first $k-1$ do not, in other words, the height of the $k$th person is $>180$cm while the heights of the first $k-1$ people are $<180$. Assuming independence of the people coming through the door we get:

$$P[Y=k]=P[X>180]\left ( P[X<180] \right )^{k-1}$$

This can be computed easily from the fact that $X$ is normal. To calculate the expected value of $Y$ you should find the series:

$$E(X)=\sum_{k=1}^{\infty}kP[Y=k]$$

Remark: We must not forget the case that no people ever bend when coming through the door, i.e."Y=$\infty$". However it is easy to see that this has probability $0$.

Also the inequalities $X>180$ can be replaced by $X≥180$ due to the fact that $X$ is continuous.