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 :

Answer:

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

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

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 [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

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

This means that [tex]p = 0.1[/tex]

The particular bag we examine contains 320 M&Ms.

This means that [tex]n = 320[/tex]

Mean and standard deviation:

[tex]\mu = p = 0.1[/tex]

[tex]s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.1*0.9}{320}} = 0.0168[/tex]

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

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

By the Central Limit Theorem

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

[tex]Z = \frac{0.12 - 0.1}{0.0168}[/tex]

[tex]Z = 1.19[/tex]

[tex]Z = 1.19[/tex] 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.