Question

A candy company claims that 31% of the candies are orange. We randomly sample 100 candies. We wish to know the probability our sample is 29% orange candies or less. Round answers to 4 decimal places and answer numerically (0.31, not 31%)

Find the probability our sample is 29% orange candies or less P(x<0.29)

Answer 1

Using the normal distribution and the central limit theorem, it is found that:

P(x < 0.29) = 0.2981.

Normal Probability Distribution

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

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

• 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 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:

• 31% of the candies are orange, hence p = 0.31.

• We randomly sample 100 candies, hence n = 100.

Then, the mean and the standard error are given by:

$\mu = p = 0.31$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.31(0.69)}{100}} = 0.04625$

The probability our sample is 29% orange candies or less is the p-value of Z when X = 0.29, hence:

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

By the Central Limit Theorem

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

$Z = \frac{0.29 - 0.31}{0.04625}$

$Z = -0.43$

$Z = -0.43$ has a p-value of 0.2981.

Hence:
P(x < 0.29) = 0.2981.

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