Question

Mars Candy claims that 10% of all M&Ms are blue. Assume that the M&Ms in a 1 pound bag form an SRS from

all M&Ms. The particular bag we examine contains 320 M&Ms. What is the probability that the bag we selected
contains more than 12% blue M&Ms?

Answer 1

0.1170 = 11.70% probability that the bag we selected contains more than 12% blue M&Ms.

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 zscore of a measure X is given by:

`Z = (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 pvalue, 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 = (\sigma)/(√(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{(p(1-p))/(n)}`

Mars Candy claims that 10% of all M&Ms are blue.

This means that
`p = 0.1`

The particular bag we examine contains 320 M&Ms.

This means that
`n = 320`

Mean and standard deviation:

`\mu = p = 0.1`

`s = \sqrt{(p(1-p))/(n)} = \sqrt{(0.1*0.9)/(320)} = 0.0168`

What is the probability that the bag we selected contains more than 12% blue M&Ms?

This is 1 subtracted by the pvalue of Z when X = 0.12. So

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (0.12 - 0.1)/(0.0168)`

`Z = 1.19`

`Z = 1.19` has a pvalue of 0.8830

1 - 0.8830 = 0.1170

0.1170 = 11.70% probability that the bag we selected contains more than 12% blue M&Ms.