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Sagot :
Using the normal distribution, it is found that there is a 0.8413 probability that a randomly chosen box of healthy oats weighs more than 8.8 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.
In this problem:
- The mean is of 8.9 oz, hence [tex]\mu = 8.9[/tex].
- The standard deviation is of 0.1 oz, hence [tex]\sigma = 0.1[/tex].
The probability that a randomly chosen box of healthy oats weighs more than 8.8 oz is 1 subtracted by the p-value of Z when X = 8.8, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{8.8 - 8.9}{0.1}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a p-value of 0.1587.
1 - 0.1587 = 0.8413.
0.8413 probability that a randomly chosen box of healthy oats weighs more than 8.8 oz.
To learn more about the normal distribution, you can take a look at https://brainly.com/question/24663213
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