The arc electronic company had an income of 90 million dollars last year. suppose the mean income of firms in the same industry as arc for a year is 75 million dollars with a standard deviation of 11 million dollars. 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? round your answer to four decimal places.

0.0869

Step-by-step explanation:

The arc electronic company had an income of 90 million dollars last year.

Mean(μ) = 75 million dollars

Standard deviation (σ) = 11 million dollars

Probability that the randomly selected will earn more than arc did last year = Pr(x>90)

Using normal distribution,

Z = (x - μ) / σ

Z = (90 - 75) / 11

Z = 15/11

Z = 1.36

From the normal distribution table, 1.36 = 0.4131

Φ(z) = 0.4131

Recall that when Z is positive, Pr(x>a) = 0.5 - Φ(z)

= 0.5 - 0.4131

= 0.0869

option: d is correct

step-by-step explanation: