,does a polygon usually have more sides or angles.explain

A building has n floors numbered 1,2,...,n, plus a ground floor g. at the ground floor, m people get on the elevator together, a

fomenos

Let $X_i$ be the random variable indicating whether the elevator does not stop at floor $i$ , with

$X_i=\begin{cases}1&\text{if the elevator does not stop at floor }i\\0&\text{otherwise}\end{cases}$

Let $Y$ be the random variable representing the number of floors at which the elevator does not stop. Then

$Y=X_1+X_2+\cdots+X_{n-1}+X_n$

We want to find $\mathrm{Var}(Y)$ . By definition,

$\mathrm{Var}(Y)=\mathbb E[(Y-\mathbb E[Y])^2]=\mathbb E[Y^2]-\mathbb E[Y]^2$

As stated in the question, there is a $\dfrac1n$ probability that any one person will get off at floor $n$ (here, $n$ refers to any of the $n$ total floors, not just the top floor). Then the probability that a person will not get off at floor $n$ is $1-\dfrac1n$ . There are $m$ people in the elevator, so the probability that not a single one gets off at floor $n$ is $\left(1-\dfrac1n\right)^m$ .

So,

$\mathbb P(X_i=x)\begin{cases}\left(1-\dfrac1n\right)^m&\text{for }x=1\\\\1-\left(1-\dfrac1n\right)^m&\text{for }x=0\end{cases}$

which means

$\mathbb E[Y]=\mathbb E\left[\displaystyle\sum_{i=1}^nX_i\right]=\displaystyle\sum_{i=1}^n\mathbb E[X_i]=\sum_{i=1}^n\left(1\cdot\left(1-\dfrac1n\right)^m+0\cdot\left(1-\left(1-\dfrac1n\right)^m\right)$
$\implies\mathbb E[Y]=n\left(1-\dfrac1n\right)^m$

and

$\mathbb E[Y^2]=\mathbb E\left[\left(\displaystyle\sum_{i=1}^n{X_i}\right)^2\right]=\mathbb E\left[\displaystyle\sum_{i=1}^n{X_i}^2+2\sum_{1\le i$

Computing $\mathbb E[{X_i}^2]$ is trivial since it's the same as $\mathbb E[X_i]$ . (Do you see why?)

Next, we want to find the expected value of the following random variable, when $i\neq j$ :

$X_iX_j=\begin{cases}1&\text{if }X_i=1\text{ and }X_j=1\\0&\text{otherwise}\end{cases}$

If $X_iX_j=0$ , we don't care; when we compute $\mathbb E[X_iX_j]$ , the contributing terms will vanish. We only want to see what happens when both floors are not visited.

$\mathbb P(X_iX_j=1)=\left(1-\dfrac2n\right)^m$
$\implies\mathbb E[X_iX_j]=\left(1-\dfrac2n\right)^m$
$\implies2\displaystyle\sum_{1\le i$

where we multiply by $n(n-1)$ because that's how many ways there are of choosing indices $i,j$ for $X_iX_j$ such that $1\le i$ .

So,

$\mathrm{Var}[Y]=n\left(1-\dfrac1n\right)^m+2n(n-1)\left(1-\dfrac2n\right)^m-n^2\left(1-\dfrac1n\right)^{2m}$

4 0

3 years ago

Out of 78 students 51 got 100% what percent did 27 students get

Nastasia [14]

There really is no way of telling. If the 51 got 100, then the rest could have gotten anything. If there is no graph, or data, or anything else to support the question, that means the 27 could have gotten any mark.

6 0

3 years ago

Write the statement as an algebraic expression.

Firdavs [7]

Answer:

$(c^{2}+d)+2(cd)$