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

for the first question it would be the second answer for the second question it would be the third answer and for the third question it would be the first answer

step-by-step explanation: