Answer:
0.62% probability that randomly chosen salary exceeds $40,000
Explanation:
Problems of normally distributed distributions are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2021/formulas/mathematics/college/c62rrp8olhnzeelpux1qvr89ehugd6fm1f.png)
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 question:
![\mu = 30000, \sigma = 4000](https://img.qammunity.org/2021/formulas/mathematics/college/3a81yxrq56ji36e4haett2nguae75dkgwf.png)
What is the probability that randomly chosen salary exceeds $40,000
This is 1 subtracted by the pvalue of Z when X = 40000. So
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2021/formulas/mathematics/college/c62rrp8olhnzeelpux1qvr89ehugd6fm1f.png)
![Z = (40000 - 30000)/(4000)](https://img.qammunity.org/2021/formulas/mathematics/college/85dopl03coqe3xe7qsjplzo2r09rbe9w7a.png)
![Z = 2.5](https://img.qammunity.org/2021/formulas/mathematics/college/2xbp3hjb7q49gb30l2lkn331fualfnzaky.png)
has a pvalue of 0.9938
1 - 0.9938 = 0.0062
0.62% probability that randomly chosen salary exceeds $40,000