A candy maker offers Child and Adult bags of jelly beans with different color mixes. The company claims that the Child mix has 30% red jelly beans, while the Adult mix contains 15% red jelly beans. Assume that the candy maker’s claim is true. Suppose we take a random sample of 50 jelly beans from the Child mix and a separate random sample of 100 jelly beans from the Adult mix. Let ^ p C and ^ p A be the sample proportions of red jelly beans from the Child and Adult mixes, respectively.

Find the probability that the proportion of red jelly beans in the Child sample is less than or equal to the proportion of red jelly beans in the Adult sample, assuming that the company’s claim is true.

Enter the probability here:

If the companies claim is to be true this statement would also be true. Since the child bag is halved in the amount of the adults, the percentage would double. There are both 15 red jelly beans in each bag. 15% of 100 is 15. 30% of 50 is 15.

Answer Link

joaobezerra joaobezerra · Answer 2 · 2022-01-14T11:50:53+01:00

Using the normal distribution and the central limit theorem, it is found that there is a 0.0212 = 2.12% probability that the proportion of red jelly beans in the Child sample is less than or equal to the proportion of red jelly beans in the Adult sample.

Normal Probability Distribution

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

It 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, which is the percentile of X.

By the Central Limit Theorem, for a proportion p in a sample of size n, the mean is [tex]\mu = p[/tex] while the standard error is of [tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex]

When two normal variables are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

For the Child mix, we have that:

[tex]p_C = 0.3, s_C = \sqrt{\frac{0.3(0.7)}{50}} = 0.0648[/tex]

For the Adult mix, we have that:

[tex]p_A = 0.15, s_A = \sqrt{\frac{0.15(0.85)}{100}} = 0.0357[/tex]

For the distribution of differences, we have that:

[tex]\mu = p_C - p_A = 0.3 - 0.15 = 0.15[/tex]

[tex]s = \sqrt{s_C^2 + s_A^2} = \sqrt{0.0648^2 + 0.0357^2} = 0.074[/tex]

The probability that the proportion of red jelly beans in the Child sample is less than or equal to the proportion of red jelly beans in the Adult sample is the p-value of Z when X = 0, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem:

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{0 - 0.15}{0.074}[/tex]

[tex]Z = -2.03[/tex]

[tex]Z = -2.03[/tex] has a p-value of 0.0212.

0.0212 = 2.12% probability that the proportion of red jelly beans in the Child sample is less than or equal to the proportion of red jelly beans in the Adult sample.

To learn more about the normal distribution and the central limit theorem, you can take a look at https://brainly.com/question/24663213