At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
Using the normal distribution and the central limit theorem, it is found that there is a:
a) 0.2902 = 29.02% probability that the amount of granola in a box selected at random will be between 13.9 and 14.2 oz.
b) 0.766 = 76.6% probability that the sample mean of granola in a random sample of 12 boxes will be between 13.9 and 14.2 oz.
Normal Probability Distribution
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It 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, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this problem:
- The mean is [tex]\mu = 14[/tex].
- The standard deviation is [tex]\sigma = 0.4[/tex].
Item a:
The probability is the p-value of Z when X = 14.2 subtracted by the p-value of Z when X = 13.9, hence:
X = 14.2:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{14.2 - 14}{0.4}[/tex]
[tex]Z = 0.5[/tex]
[tex]Z = 0.5[/tex] has a p-value of 0.6915.
X = 13.9:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{13.9 - 14}{0.4}[/tex]
[tex]Z = -0.25[/tex]
[tex]Z = -0.25[/tex] has a p-value of 0.4013.
0.6915 - 0.4013 = 0.2902.
0.2902 = 29.02% probability that the amount of granola in a box selected at random will be between 13.9 and 14.2 oz.
Item b:
Sample of 12, hence n = 12 and [tex]s = \frac{0.4}{\sqrt{12}} = 0.1155[/tex].
X = 14.2:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{14.2 - 14}{0.1155}[/tex]
[tex]Z = 1.73[/tex]
[tex]Z = 1.73[/tex] has a p-value of 0.9582.
X = 13.9:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{13.9 - 14}{0.1155}[/tex]
[tex]Z = -0.87[/tex]
[tex]Z = -0.87[/tex] has a p-value of 0.1922.
0.9582 - 0.1922 = 0.766.
0.766 = 76.6% probability that the sample mean of granola in a random sample of 12 boxes will be between 13.9 and 14.2 oz.
To learn more about the normal distribution and the central limit theorem, you can check https://brainly.com/question/24663213
We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
