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Sagot :
Answer:
The value of 14.36 lb separates the top 2% of the weekly discarded paper weights in this area.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set 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]
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 p-value, we get the probability that the value of the measure is greater than X.
Normally distributed with the average of 7.95 lb and the standard deviation of 3.12 lb.
This means that [tex]\mu = 7.95, \sigma = 3.12[/tex]
Which value separates the top 2% of the weekly discarded paper weights in this area?
This is the 100 - 2 = 98th percentile, which is X when Z has a pvalue of 0.98. So X when Z = 2.054.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]2.054 = \frac{X - 7.95}{3.12}[/tex]
[tex]X - 7.95 = 2.054*3.12[/tex]
[tex]X = 14.36[/tex]
The value of 14.36 lb separates the top 2% of the weekly discarded paper weights in this area.
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