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Sagot :
Using the normal distribution, it is found that:
- The worker's score is at the 86.4th percentile.
- 42.2% of people in this profession earn annual salaries between $54,000 and $57,000.
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.
Question 1:
1.1 standard deviations above the mean, hence, a z-score of Z = 1.1.
- Looking at the z-table, z = 1.1 has a p-value of 0.864, hence, the worker's score is at the 86.4th percentile.
Question 2:
- Mean of 54800, hence [tex]\mu = 54800[/tex]
- Standard deviation of 2600, hence [tex]\sigma = 2600[/tex].
The proportion is the p-value of Z when X = 57000 subtracted by the p-value of Z when X = 54000, then:
X = 57000:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{57000 - 54800}{2600}[/tex]
[tex]Z = 0.846[/tex]
[tex]Z = 0.846[/tex] has a p-value of 0.801.
X = 54000:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{54000 - 54800}{2600}[/tex]
[tex]Z = -0.307[/tex]
[tex]Z = -0.307[/tex] has a p-value of 0.379.
0.801 - 0.379 = 0.422
0.422 x 100% = 42.2%
42.2% of people in this profession earn annual salaries between $54,000 and $57,000.
A similar problem is given at https://brainly.com/question/24663213
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