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The weight of granola in the box is normally distributed with mean 14 oz. and standard deviation
0.4 oz.
a) What is the probability that the amount,×, of granola in a box selected at random will be
between 13.9 and 14.2 oz.
b) What is the probability that the sample mean of granola, I in a random sample of 12 boxes will
be between 13.9 and 14.2 oz?


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