Question

Approximately 70% of statistics students do their homework in time for it to be collected and graded. Each student does homework independently. In a statistics class of 50 students, what is the probability that 37 will do their homework on time

Answer 1

Answer:

0.105 = 10.5% probability that 37 will do their homework on time.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they do their homework on time, or they do not. Each student does homework independently, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Approximately 70% of statistics students do their homework in time for it to be collected and graded.

This means that $p = 0.7$

Statistics class of 50 students

This means that $n = 50$

What is the probability that 37 will do their homework on time?

This is P(X = 37). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 37) = C_{50,37}.(0.7)^{37}.(0.3)^{13} = 0.105$

0.105 = 10.5% probability that 37 will do their homework on time.