Respuesta :
Answer:
23.89% probability that a randomly selected firm will earn more than Arc did last year
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 = 80, \sigma = 7[/tex]
If incomes for this industry are distributed normally, what is the probability that a randomly selected firm will earn more than Arc did last year
This is 1 subtracted by the pvalue of Z when X = 85. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{85 - 80}{7}[/tex]
[tex]Z = 0.71[/tex]
[tex]Z = 0.71[/tex] has a pvalue of 0.7611
1 - 0.7611 = 0.2389
23.89% probability that a randomly selected firm will earn more than Arc did last year
