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A local shipping company, Company #1, specializes in shipping small packages only. Employees weigh outgoing packages, and find that the weights are approximately Normally distributed with a mean of 2.3 pounds and a standard deviation of 1.2 pounds. Another shipping company, Company #2, specializes in large pallets of containers. For this shipping company, the weights of pallets are also approximately Normally distributed with a mean of 266 pounds and a standard deviation of 10.7 pounds.

(a) What is the 80th percentile of small package weights at Company #1?

(b) At Company A, packages weighing more than 5 pounds require additional fees. What proportion of small packages at this shipping company require additional fees?

(c) Which is heavier relative to other items shipped at the same company, an 8.1-pound package at Company A or a 265-pound pallet at Company B?

(d)Show that a pallet weighing 295 pounds would be an outlier at Company B.

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.

A similar problem is given at https://brainly.com/question/22934264

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