Respuesta :
Answer:
8.85% probability that her revenue will exceed $600,000.
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 = 425000, \sigma = 130000[/tex]
What is the probability that her revenue will exceed $600,000?
This is 1 subtracted by the pvalue of Z when X = 600000.
So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{600000 - 425000}{130000}[/tex]
[tex]Z = 1.35[/tex]
[tex]Z = 1.35[/tex] has a pvalue of 0.9115.
So there is a 1-0.9115 = 0.0885 = 8.85% probability that her revenue will exceed $600,000.
