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Answer :
Answer:
To be in the top 5%, a person must have consumed at least 149.25 pounds of fruit.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 99.9, \sigma = 30[/tex]
How much fruit did a person have to consume to be in the top 5%
At least X pounds, in which X is found when Z has a pvalue of 1-0.05 = 0.95. So it is X when Z = 1.645.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.645 = \frac{X - 99.9}{30}[/tex]
[tex]X - 99.9 = 30*1.645[/tex]
[tex]X = 149.25[/tex]
To be in the top 5%, a person must have consumed at least 149.25 pounds of fruit.
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