### Probability and Statistics with SkittlesConsider the number of green Skittles in a package of Skittles. Suppose 29% of packages contain 3 green Skittles, 26% contain 4 green Skittles, and 45% contain 5 green Skittles.Let \( X \) be the number of green Skittles in a randomly chosen pack of Skittles. In Midterm 1, we computed the expected value, \( E(X) \), to be 4.16 and the standard deviation to be 0.85. Suppose you choose 100 packs of Skittles at random. What is the chance that the average number of green Skittles (in the 100 packs) is less than 4? Answer to two decimal places.#### CalculationUse the central limit theorem to approximate this probability.Let \( \bar{X} \) be the sample mean of the number of green Skittles in 100 packs. Then,- The mean of \( \bar{X} \), \( E(\bar{X}) = E(X) = 4.16 \).- The standard deviation of \( \bar{X} \), \( \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{0.85}{\sqrt{100}} = \frac{0.85}{10} = 0.085 \).We need to find \( P(\bar{X} < 4) \). Standardize the variable:\[ Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} = \frac{4 - 4.16}{0.085} \]Calculate the z-value:\[ Z = \frac{-0.16}{0.085} \approx -1.88 \]Using the standard normal distribution table, find the probability corresponding to \( Z = -1.88 \), which is approximately 0.03.Thus, the probability that the average number of green Skittles in the 100 packs is less than 4 is approximately 0.03, or 3%.

Z-score: The z-score value is a numerical measurement that describes a value of the relationship to…

Consider the number of green Skittles in a package of Skittles. Suppose 29% of packages contain 3 green Skittles, 26% contain 4 green Skittles, and 45% contain 5 green Skittles. Let X be the number of green Skittles in a randomly chosen pack of Skittles. In Midterm 1 we computed the expected value, E(X), to be 4.16 and the standard deviation to be 0.85. Suppose you choose 100 packs of Skittles at random. What is the chance that the average number of green Skittles (in the 100 packs) is less than 4? Answer to two decimal places.

Consider the number of green Skittles in a package of Skittles. Suppose 29% of packages contain 3 green Skittles, 26% contain 4 green Skittles, and 45% contain 5 green Skittles. Let X be the number of green Skittles in a randomly chosen pack of Skittles. In Midterm 1 we computed the expected value, E(X), to be 4.16 and the standard deviation to be 0.85. Suppose you choose 100 packs of Skittles at random. What is the chance that the average number of green Skittles (in the 100 packs) is less than 4? Answer to two decimal places.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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