Respuesta :
Answer:
0.9772 = 97.72% probability that a randomly selected firm will earn more than Arc did last year
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions 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.
Suppose the mean income of firms in the same industry as Arc for a year is 90 million dollars with a standard deviation of 7 million dollars.
This means that [tex]\mu = 90, \sigma = 7[/tex]
What is the probability that a randomly selected firm will earn more than Arc did last year?
Arc earned 76 million, so this is 1 subtracted by the pvalue of Z when X = 76.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{76 - 90}{7}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] has a pvalue of 0.0228
1 - 0.0228 = 0.9772
0.9772 = 97.72% probability that a randomly selected firm will earn more than Arc did last year
