Find expert answers and community support for all your questions on IDNLearn.com. Ask your questions and receive prompt, detailed answers from our experienced and knowledgeable community members.
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
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. IDNLearn.com has the solutions you’re looking for. Thanks for visiting, and see you next time for more reliable information.
