Respuesta :
Answer:
d. 76.98%
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 = 40000, \sigma = 5000[/tex]
What percentage of MBA's will have starting salaries of $34,000 to $46,000?
This is the pvalue of Z when X = 46000 subtracted by the pvalue of Z when X = 34000. So
X = 46000
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{46000 - 40000}{5000}[/tex]
[tex]Z = 1.2[/tex]
[tex]Z = 1.2[/tex] has a pvalue of 0.8849
X = 34000
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{34000 - 40000}{5000}[/tex]
[tex]Z = -1.2[/tex]
[tex]Z = -1.2[/tex] has a pvalue of 0.1151
0.8849 - 0.1151 = 0.7698
So the correct answer is:
d. 76.98%
