Mathematics, 08.03.2021 14:00 calvinsanchez8429

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?

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Answer from: missalawode28

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 $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

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 $s = \frac{\sigma}{\sqrt{n}}$ .

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 $s = \sqrt{\frac{p(1-p)}{n}}$

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:

$\mu = p = 0.1$

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

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

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

By the Central Limit Theorem

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

$Z = \frac{0.12 - 0.1}{0.0168}$

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.

Answer from: Quest

1+1=22×1=22-1=11-1=1duh

Answer from: Quest

inverse operation of addition

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

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