Question

A binomial experiment is given. Determine whether you can use a normal distribution to approximate the binomial distribution. If you can, find the mean and standard deviation. If you cannot, explain why. A survey of U.S. adults found that 37% have been to court. You randomly select 30 U.S. adults and ask them whether they have been to court.

Answer 1

Answer:

Since both $np \geq 10$ and $n(1-p) \geq 10$, you can use a normal distribution to approximate the binomial distribution.

The mean is of 11.1 and the standard deviation is of 2.64.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

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

Using the normal distribution to approximate the binomial distribution.

This is possible if:

$np \geq 10, n(1-p) \geq 10$

A survey of U.S. adults found that 37% have been to court. You randomly select 30 U.S.

This means that $p = 0.37, n = 30$

Test if it is possible:

$np = 30*0.37 = 11.1$

$n(1-p) = 30*0.63 = 18.9$

Since both $np \geq 10$ and $n(1-p) \geq 10$, you can use a normal distribution to approximate the binomial distribution.

Mean and standard deviation:

$E(X) = np = 30*0.37 = 11.1$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{30*0.37*0.63} = 2.64$

The mean is of 11.1 and the standard deviation is of 2.64.