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.​
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joaobezerra 7 months ago
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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 and standard deviation is given by:

• 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 and standard deviation , as long as and .

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

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:

By the Central Limit Theorem:

Z = 0.5

Z = 0.5 has a p-value of 0.6915.

X = 0.18:

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://brainacademy.pro/question/24663213