Question

The Arc Electronic Company had an income of 72 million dollars last year. Suppose the mean income of firms in the same industry as Arc for a year is 85 million dollars with a standard deviation of 9 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.

Answer 1

Solution

Let X = income (million dollars) of firms in the same industry as Arc for a year. Then, we are given: X ~ N(85, 9²) ….. (1)

Back-up Theory

If a random variable X ~ N(µ, σ²), i.e., X has Normal Distribution with mean µ and variance σ², then, Z = (X - µ)/σ ~ N(0, 1), i.e., Standard Normal Distribution and hence

P(X ≤ or ≥ t) = P[{(X - µ)/σ} ≤ or ≥ {(t - µ)/σ}] = P[Z ≤ or ≥ {(t - µ)/σ}] .………….....................................................……...…(2)

Probability values for the Standard Normal Variable, Z, can be directly read off from Standard Normal Tables ............. (3a)

or can be found using Excel Function: Statistical, NORMSDIST(z) which gives P(Z ≤ z) .............................................…(3b)

Now to work out the solution,

Given, “The Arc Electronic Company had an income of 72 million dollars last year”, probability that a randomly selected firm will earn more than Arc did last year

= P(X > 72)

= P[Z > {(72 - 85)/9}] [vide (2) and (1)]

= P(Z > 0.1529)

= 0.4392 [vide (3b)] Answer

DONE