Question

According to the Mars company, packages of milk chocolate M&Ms contain 20% orange candies. Find the probability that in a random sample of 100 M&M candies, there are between 18% and 22% orange candies.​

Answer 1

Using the normal distribution, it is found that there is a 0.383 = 38.3% probability that the sample proportion is between 18% and 22%.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

• By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$, as long as $np \geq 10$ and $n(1 - p) \geq 10$.

In this problem, the proportion and the sample size are of p = 0.2 and n = 100, respectively, hence:

$\mu = p = 0.2$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.2(0.8)}{100}} = 0.04$

The probability that in a random sample of 100 M&M candies, there are between 18% and 22% orange candies is the p-value of Z when X = 0.22 subtracted by the p-value of Z when X = 0.18, hence:

X = 0.22:

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

By the Central Limit Theorem:

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

$Z = \frac{0.22 - 0.2}{0.04}$

Z = 0.5

Z = 0.5 has a p-value of 0.6915.

X = 0.18:

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

$Z = \frac{0.18 - 0.2}{0.04}$

Z = -0.5

Z = -0.5 has a p-value of 0.3085.

0.6915 - 0.3085 = 0.383.

0.383 = 38.3% probability that the sample proportion is between 18% and 22%.

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