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.

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### Probability and Statistics with Skittles

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.

#### Calculation
Use 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%.
Transcribed Image Text:### Probability and Statistics with Skittles 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. #### Calculation Use 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%.
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