Question

question 10
1 pts
it has been found that 88% of umsl students live off campus. suppose 50 students are selected at random and that the number of students out of those 50 living off campus follows a binomial distribution. what is the standard deviation for the number them who live off campus?
○ 5.28
○ 2.449
○ 6.633
○ 44
○ 12
○ 2.298

Explanation:

Step1: Identify binomial parameters

For a binomial distribution, \( n = 50 \) (number of trials), \( p = 0.88 \) (probability of success, living off - campus), and \( q=1 - p=1 - 0.88 = 0.12 \) (probability of failure).

Step2: Use standard deviation formula for binomial

The formula for the standard deviation (\(\sigma\)) of a binomial distribution is \(\sigma=\sqrt{npq}\).

Substitute \( n = 50 \), \( p = 0.88 \), and \( q = 0.12 \) into the formula:

First, calculate \( npq \):
\( npq=50\times0.88\times0.12 \)
\( 50\times0.88 = 44 \)
\( 44\times0.12=5.28 \)

Then, take the square root of \( npq \):
\(\sigma=\sqrt{5.28}\approx2.298\)

Answer:

2.298