### 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.

College Algebra

10th Edition

ISBN:9781337282291

Author:Ron Larson

Publisher:Ron Larson

Chapter8: Sequences, Series,and Probability

Section8.7: Probability

Problem 4ECP: Show that the probability of drawing a club at random from a standard deck of 52 playing cards is...

See similar textbooks

Related questions

Q: In a Z-test, the number of degrees of freedom is A none of these B 1 C) n-1 D n-2

A: AS PER GUIDELINE EXPERT HAVE TO ANSWER FIRST QUESTION ONLY. SOJCE I HAD DONE ALL FOR YOU. AND ALL…

Q: Let x be a random variable that represents the pH of arterial plasma (i.e., acidity of the blood).…

A: The sample size n nis 36.

Q: Let x be a random variable that represents the pH of arterial plasma (i.e., acidity of the blood).…

A: Determine the level of significance. Significance level: In order to decide how small the p-value…

Q: A person's blood glucose level and diabetes are closely related. Let x be a random variable measured…

A: GivenMean(μ)=86standard deviation(σ)=23

Q: Let x be a random variable that represents the pH of arterial plasma (i.e., acidity of the blood).…

A: Givenμ=7.4sample size (n)=41x=8.4s=2.7

Q: For the majority of maths tests taken at a school, it is shown that the test score is approximaly…

A: Dear Student, We are allowed to solve only the first 3 subparts. Kindly repost the remaining part.…

Q: A sample of size 52 will be drawn from a population with a mean of 18 and a standard deviation of…

Q: If n=7 (days in a week) and p=0.45 (the probability that the food truck operates on a single day),…

A: Given information- Sample size, n = 7 The probability that the food truck operates on a single day…

Q: O A. 0.8615 OB. 0.8819 O C. 0.0062 OD. 0.9938

A: Let X be the random variable from normal distribution with mean (μ) = 124, standard deviation (σ) =…

Q: The capacity of an elevator is 15 people or 2400 pounds. The capacity will be exceeded if 15 people…

A: a. The Z-score of a random variable X is defined as follows: Z = (X – µ)/σ. Here, µ and σ are the…

Concept explainers

Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.

Question

### 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%.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Point Estimation, Limit Theorems, Approximations, and Bounds$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.

Similar questions

A manufacturer has determined that a machine averages one faulty unit for every 500 it produces. What is the probability that an order of 300 units will have one or more faulty units?
If a binomial experiment has probability p success, then the probability of failure is ____________________. The probability of getting exactly r successes in n trials of this experiment is C(_________, _________)p (1p)
What term is used to express the likelihood of an event occurring? Are there restrictions on its values? If so, what are they? If not, explain.
Show that the probability of drawing a club at random from a standard deck of 52 playing cards is the same as the probability of drawing the ace of hearts at random from a set of four cards consisting of the aces of hearts, diamonds, clubs, and spades.