Respuesta :
Answer:
a) 0.219
b) 0.02
Step-by-step explanation:
Problems of normally distributed samples can be 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 = 43000, \sigma = 18000[/tex]
a. What percent of high school teachers make between $40,000 and $50,000?
This is the pvalue of Z when X = 50,000 subtracted by the pvalue of Z when X = 40,000.
So
X = 50,000
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{50000 - 43000}{18000}[/tex]
[tex]Z = 0.39[/tex]
[tex]Z = 0.39[/tex] has a pvalue of 0.6517.
X = 40,000
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{40000 - 43000}{18000}[/tex]
[tex]Z = -0.17[/tex]
[tex]Z = -0.17[/tex] has a pvalue of 0.4325
So 0.6517 - 0.4325 = 0.219 = 21.9% of high school teachers make between $40,000 and $50,000.
b. What percent of high school teachers make more than $80,000?
This is 1 subtracted by the pvalue of Z when X = 80000. SO
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{80000 - 43000}{18000}[/tex]
[tex]Z = 2.06[/tex]
[tex]Z = 2.06[/tex] has a pvalue of 0.9803.
So 1-0.9803 = 0.0197 = 0.02 = 2% of high school teachers make more than $80,000.
