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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
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