Question

According to national data, about 14% of American college students earn a graduate degree. Using this estimate, what is the probability that exactly 26 undergraduates in a random sample of 200 students will earn a college degree? Hint: Use the normal approximation to the binomial distribution, where p = 0.14 and q = 0.86. (Round your answer to four decimal places.)

Answer 1

Answer:

$X \sim Binom (n =200, p =0.14)$

Using the probability mass function we can find the probability required

$P(X=26) = (200C26) (0.14)^{26} (1-0.14)^{200-26} = 0.0768$

We need to check the conditions in order to use the normal approximation.

We assume independence in the events

$np = 200*0.14 =28 \geq 10$

$n(1-p)=200*(1-0.14)=172 \geq 10$

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

$E(X)=np=200*0.14=28$

$\sigma=\sqrt{np(1-p)}=\sqrt{200*0.14(1-0.14)}=4.907$

So then we can approximate the normal with $X \sim N(\mu =28, \sigma =4.907)$

But for this case since we have a continuous distribution with this approximation we can't find $P(X=26)$ since the are below a point is 0, w can find $P(X>26)$ or $P(X<26)$ but not $P=26$ using the normal approximation.

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".  

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

$X \sim Binom (n ,p)$

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:

$nCx=\frac{n!}{(n-x)! x!}$  

Solution to the problem

For this case we have the following distirbuion for X the number of  undergraduates

$X \sim Binom (n =200, p =0.14)$

Using the probability mass function we can find the probability required

$P(X=26) = (200C26) (0.14)^{26} (1-0.14)^{200-26} = 0.0768$

We need to check the conditions in order to use the normal approximation.

We assume independence in the events

$np = 200*0.14 =28 \geq 10$

$n(1-p)=200*(1-0.14)=172 \geq 10$

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

$E(X)=np=200*0.14=28$

$\sigma=\sqrt{np(1-p)}=\sqrt{200*0.14(1-0.14)}=4.907$

So then we can approximate the normal with $X \sim N(\mu =28, \sigma =4.907)$

But for this case since we have a continuous distribution with this approximation we can't find $P(X=26)$ since the are below a point is 0, w can find $P(X>26)$ or $P(X<26)$ but not $P=26$ using the normal approximation.