Answer:
21.77% probability that a randomly selected teacher earns more than $525 a week
Step-by-step explanation:
Problems of normally distributed samples are 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 = 490, \sigma = 45[/tex]
What is the probability that a randomly selected teacher earns more than $525 a week?
This is 1 subtracted by the pvalue of Z when X = 525. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{525 - 490}{45}[/tex]
[tex]Z = 0.78[/tex]
[tex]Z = 0.78[/tex] has a pvalue of 0.7823
1 - 0.7823 = 0.2177
21.77% probability that a randomly selected teacher earns more than $525 a week
There is a probability of 21.77% that selected teacher earns more than $525 a week.
Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
[tex]z=\frac{x-\mu}{\sigma} \\\\wher\ \mu=mean,x=raw\ score,\sigma=standard\ deviation[/tex]
μ = 490, σ = 45:
For x > 525:
[tex]z=\frac{525-490}{45}=0.78[/tex]
From the normal distribution table, P(x > 525) = P(z > 0.78) = 1 - P(z < 0.78) = 1 - 0.7823 = 21.77%
There is a probability of 21.77% that selected teacher earns more than $525 a week.
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