Question

At a new exhibit in the Museum of Science, people are asked to choose between 94 or 220 random draws from a machine. The machine is known to have 99 green balls and 78 red balls. After each draw, the color of the ball is noted and the ball is put back for the next draw. You win a prize if more than 61% of the draws result in a green ball. a. Calculate the probability of getting more than 61% green balls.

Answer 1

Answer:

0.0869 = 8.69% probability of getting more than 61% green balls.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

The machine is known to have 99 green balls and 78 red balls.

This means that $p = \frac{99}{99+78} = 0.5593$

Mean and standard deviation:

$\mu = p = 0.5593$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5593*0.4407}{99+78}} = 0.0373$

a. Calculate the probability of getting more than 61% green balls.

This is 1 subtracted by the pvalue of Z when X = 0.61. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.61 - 0.5593}{0.0373}$

$Z = 1.36$

$Z = 1.36$ has a pvalue of 0.9131

1 - 0.9131 = 0.0869

0.0869 = 8.69% probability of getting more than 61% green balls.