 # At a certain college, 6% of all students come from outside the United States. Incoming students there are assigned at random to freshman dorms, where students live in residential clusters of $40$ freshmen sharing a common lounge area.

• How many international students would you expect to find in a typical cluster?

• With what standard deviation?

This question aims to find the expected number of international students in a typical cluster along with their standard deviation.

Take into consideration what a random variable is: a collection of numerical values resulting from a random process. The weighted mean of independent occurrences is used to get the expected values. In general, it employs probability to predict the long-term occurrences required. The standard deviation is a measure of how far a set of numerical values shifts away from its mean.

The international students are the random variable (number of successes) in this question, and the proportion of international students is the chance of success.

Each student might be either an international student or a permanent resident of the United States. The likelihood of a foreign student is irrespective of the probability of other students in this context; hence we should utilize the Binomial distribution.

Let $X$ denote the number of successes, $n$ denote the number of trials and $p$ represent the success probability. The probability of failure will then be $1-p$.

The expected value of $X$ is specified as

$\mu=E(X)=np$

And the standard deviation is

$\sigma=\sqrt{V(X)}=\sqrt{npq}=\sqrt{np(1-p)}$

Where variance is $V(X)$.

Given the problem stated above:

The success probability is international students. As there are $6\%$ of international students so,

$p=6\%=0.06$

Also, we have samples of $40$ students, therefore,

$n=40$

## Numerical Results

$\mu=E(X)=np=(40)(0.06)=2.4$

$\sigma=\sqrt{np(1-p)}=\sqrt{(40)(0.06)(1-0.06)}=\sqrt{(40)(0.06)(0.94)}=1.5$

Hence, $2.4$ international students are expected in a typical cluster having the standard deviation of $1.5$ students.

## Alternative solution

The success probability   $=p$

Then probability of failure $=q=1-p$

As $p=0.06$ so $q=1-0.06=0.94$

$\mu=E(X)=np=(40)(0.06)=2.4$

And the standard deviation is

$\sigma= \sqrt{npq}= \sqrt{(40)(0.06)(0.94)}=1.5$

The above problem is graphically illustrated as: ## Example

A binomial trial has $60$ occurrences. The failure probability for every trial is $0.8$. Find the expected value and the variance.

Here, the number of trials $n=60$, and the failure probability $q=0.8$

It is well known that

$q=1-p$

So,

$p=1-q=1-0.8=0.2$

Hence,

$\mu=E(X)=np=(60)(0.2)=12$

$\sigma^2=npq=(60)(0.2)(0.8)=9$

So from the example, we can observe the same results when either the probability of success or failure is given.

Images/mathematical drawings are created with GeoGebra.

5/5 - (11 votes)