Question

More than 96 percent of the very largest colleges and universities (more than 15,000 total enrollments) have some online offerings. Suppose you randomly pick 13 such institutions. We are interested in the number that offer distance learning courses.

a. In words, define the random variable X.

b. List the values that X may take on.

c. Give the distribution of X. X ~ _____(_____,_____)

d. On average, how many schools would you expect to offer such courses?

e. Find the probability that at most ten offer such courses.

f. Is it more likely that 12 or that 13 will offer such courses? Use numbers to justify your answer numerically and answer in a complete sentence.

Short answer

a. X is the number of colleges and universities that offer online offerings.

b. The values that X may take on are $0,1,2,....,13.$

c. The distribution is $X~B(13,0.96)$

d. On an average the number of schools would expect to offer the courses are $12.48$

e. The probability that at most ten offer such courses is $0.0135$

f. Most likely$13$will offer such courses.

Step-by-step solution

Content Introduction

The binomial distribution determines the probability of looking at a specific quantity of a hit results in a specific quantity of trials.

Part (a) Step 1: Explanation 

We are given,

More than $96\%$of the very largest colleges and universities have some online offerings and we have randomly picked $13$such institutions.

Random variable in simple terms generally refers to variables whose values are unknown, therefore, in this case the random variable X is the number of colleges and universities that offer online offerings.

Part (b) Step 1: Explanation 

Make the list of values that you want to use X may take on.

As we can see there is an upper bound for the situation at hand, $13$then X is given by,

$X=0,1,2,....,13.$

Part (c) Step 1: Explanation 

The random variable is distributed by the data provided X will keep the track of online offerings.

According to the given information, $96\%$colleges and universities have some online offerings.

The probability distribution of binomial distribution has two parameters $n=numberoftrials$and $p=probabilityofsuccess$. The binomial distribution is of the form:

$X~B(n,p)$

Therefore, according to given information, role="math" localid="1649077939499" $n=13$is the number of colleges that offer online offerings and $$\begin{gathered} p=\frac{96}{100} \\ p=0.96 \end{gathered}$$

Hence,$X~B(13,0.96)$

Part (d) Step 1: Explanation 

The expected binomial distribution is calculated as:

$\mu =np$where, $\mu$is the average number of schools that would offer such courses, $n=13$is the number of colleges offer online offering and $p=0.96$is the probability of success.

$$\begin{gathered} \mu =13\times 0.96 \\ \mu =12.48 \end{gathered}$$

Part (e) Step 1: Explanation 

The probability that at most ten offer such courses is $$\begin{gathered} {=}_{13}{C}_{10}{(0.96)}^{10}{(0.04)}^3 \\ =0.0135 \end{gathered}$$

Part (f) Step 1: Explanation 

Using $TI-83$calculator we have found out that the probability of $12$offering such course and the probability $13$of offering such courses are as follow:

$$\begin{gathered} P(X=12)=0.3186 \\ P(X=13)=0.5882 \end{gathered}$$

Therefore, we can conclude from here that we are more likely to get$13.$