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Sagot :
Using the normal distribution, we have that:
a) The 80th percentile of small package weights at Company #1 is of 3.3 pounds.
b) 0.0122 = 1.22% of small packages at this shipping company require additional fees.
c) Due to the higher z-score, the 8.1 pound package at Company A is heavier relative to items shipped at the same company.
d) |Z| > 2, thus, it is an outlier.
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.
- If |Z| > 2, the measure is considered an outlier.
Item a:
- For Company 1, the mean is of 2.3 pounds, thus [tex]\mu = 2.3[/tex].
- The standard deviation is of 1.2 pounds, thus [tex]\sigma = 1.2[/tex].
- The 80th percentile is X when Z has a p-value of 0.8, so X when Z = 0.84.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.84 = \frac{X - 2.3}{1.2}[/tex]
[tex]X - 2.3 = 0.84(1.2)[/tex]
[tex]X = 3.3[/tex]
The 80th percentile of small package weights at Company #1 is of 3.3 pounds.
Item b:
This proportion is 1 subtracted by the p-value of Z when X = 5, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5 - 2.3}{1.2}[/tex]
[tex]Z = 2.25[/tex]
[tex]Z = 2.25[/tex] has a p-value of 0.9878.
1 - 0.9878 = 0.0122.
0.0122 = 1.22% of small packages at this shipping company require additional fees.
Item c:
The z-score for an 8.1 pound package at Company A is:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{8.1 - 2.3}{1.2}[/tex]
[tex]Z = 4.83[/tex]
For Company 2, we have [tex]\mu = 266, \sigma = 10.7[/tex].
For a 265 pound package, the z-score is:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{265 - 266}{10.7}[/tex]
[tex]Z = -0.09[/tex]
Due to the higher z-score, the 8.1 pound package at Company A is heavier relative to items shipped at the same company.
Item d:
295 pounds, so [tex]X = 295[/tex], and the z-score is:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{295 - 266}{10.7}[/tex]
[tex]Z = 2.71[/tex]
|Z| > 2, thus, it is an outlier.
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