Question

Use the following information to answer the next five exercises: Suppose that a group of statistics students is divided into two groups: business majors and non-business majors. There are $16$business majors in the group and seven non-business majors in the group. A random sample of nine students is taken. We are interested in the number of business majors in the sample.

On average $\left(\mu \right)$, how many would you expect to be business majors?

Short answer

In the group of statistics students, the predicted number (average) of business majors is$\mathit{\mu }=\mathbf{6}\mathbf{.}\mathbf{26}\approx \mathbf{6}$

Step-by-step solution

Formula used

The mean of a probability distribution with parameters $r,b\&n$is:

$\mu =\frac{nr}{r+b}$

Given

Already we are having$r=16,b=7$and $n=9$

Calculation

The mean is calculated by substituting $r=16,b=7$and $n=9$into the formula above:

$\mu =\frac{9^{*}16}{16+7}=6.2609$