Question

Suppose that 25 students in an AP Statistics class independently do this exercise for homework and that all of their calculators are working properly. Find the probability that at least one of them makes a Type I error.

Answer 1

Using the binomial distribution, it is found that there is a 0.999977 = 99.9977% probability that at least one of them makes a Type I error.

What is the binomial distribution formula?

The formula is:

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

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

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• There are 25 students, hence n = 25.

• 70% do not commit any error, hence 30% do and p = 0.3.

The probability that at least one commits an error is given by:

$P(X \geq 1) = 1 - P(X = 0)$

In which:

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

$P(X = 0) = C_{30,0}.(0.3)^{0}.(0.7)^{30} = 0.000023$

Then:

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.000023 = 0.999977$

There is a 0.999977 = 99.9977% probability that at least one of them makes a Type I error.

More can be learned about the binomial distribution at brainly.com/question/24863377