Question

Faith is a 95% free throw shooter. At practice, each player shoots 20 free throws. Let X = the number of free throws Faith makes out of 20 shots. (round all answers to 3 decimal places). Calculate the mean of X. Calculate the standard deviation of X. Use the binomial probability formula to find P(X = 19). Her coach says that he will let the team out of practice early if Faith makes 18 or more free throws. What is the probability that practice ends early?

Answer 1

Using the binomial distribution, it is found that:

• The mean of X is of 19.

• The standard deviation of X is of 0.975.

• P(X = 19) = 0.3774.

• There is a 0.9246 = 92.46% probability that practice ends early.

Binomial probability distribution

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

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

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.

The mean of X is:

$E(X) = np$

The standard deviation of X is:

$\sqrt{V(X)} = \sqrt{np(1 - p)}$

In this problem:

• Faith is a 95% free throw shooter, hence $p = 0.95$.

• Each player shoots 20 free throws, hence $n = 20$.

The mean and the standard deviation are:

$E(X) = np = 20(0.95) = 19$

$\sqrt{V(X)} = \sqrt{20(0.95)(0.05)} = 0.975$

P(X = 19) is given by:

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

$P(X = 19) = C_{20,19}.(0.95)^{19}.(0.05)^{1} = 0.3774$

The probability that practice ends early is:

$P(X \geq 18) = P(X = 18) + P(X = 19) + P(X = 20)$

In which:

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

$P(X = 18) = C_{20,18}.(0.95)^{18}.(0.05)^{2} = 0.1887$

$P(X = 19) = C_{20,19}.(0.95)^{19}.(0.05)^{1} = 0.3774$

$P(X = 20) = C_{20,20}.(0.95)^{20}.(0.05)^{0} = 0.3585$

Then:

$P(X \geq 18) = P(X = 18) + P(X = 19) + P(X = 20) = 0.1887 + 0.3774 + 0.3585 = 0.9246$

There is a 0.9246 = 92.46% probability that practice ends early.

You can learn more about the binomial distribution at brainly.com/question/24863377