Question

Suppose in the University of Manitoba, 40% of the students live in apartments. If 600 students are randomly selected, then the approximate probability that the number of them living in apartments will be between 200 and 400 is:a. .9267b. .9996c. .9799d. .9854e. .987

Answer 1

Answer:

"=BINOM.DIST(400,600,0.4,TRUE)-BINOM.DIST(199,600,0.4,TRUE)"

And we got:

$P(200 \leq X \leq 400)=0.999675$

The best answer for this case is:

b. .9996

The other way is using the normal approximation, but for this case they want the exact answer so is better use the binomial formula.

$\mu = np =600*0.4=240 \geq 10$

$\sigma= \sqrt{n*p(1-p)} =\sqrt{600*0.4*0.6}=12$

Conditions satisfied so the normal approximation would be: $X \sim N(\mu =240, \sigma =12)$

And we want this probability:

$P(200\leq X \leq 400) = P(\frac{200-240}{12} \leq Z \leq \frac{400-240}{12}) =P(-3.33 \leq Z \leq 13.333) = P(Z<13.33) -P(Z<-3.33) \approx 0.9996$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=600, p=0.4)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

For this case we want this probability:

$P(200 \leq X \leq 400)$

And we can use the following excel code to find this probability:

"=BINOM.DIST(400,600,0.4,TRUE)-BINOM.DIST(199,600,0.4,TRUE)"

And we got:

$P(200 \leq X \leq 400)=0.999675$

The best answer for this case is:

b. .9996

The other way is using the normal approximation, but for this case they want the exact answer so is better use the binomial formula.

$\mu = np =600*0.4=240 \geq 10$

$\sigma= \sqrt{n*p(1-p)} =\sqrt{600*0.4*0.6}=12$

Conditions satisfied so the normal approximation would be: $X \sim N(\mu =240, \sigma =12)$

And we want this probability:

$P(200\leq X \leq 400) = P(\frac{200-240}{12} \leq Z \leq \frac{400-240}{12}) =P(-3.33 \leq Z \leq 13.333) \approx 0.9996$