Suppose the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 5 million dollars. If incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn less than 100 million dollars

We are given that the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 5 million dollars. Also, incomes for the industry are distributed normally.

Let X = incomes for the industry

So, X ~ N([tex]\mu=95,\sigma^{2}=5^{2}[/tex])

Now, the z score probability distribution is given by;

Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = mean income of firms in the industry = 95 million dollars

[tex]\sigma[/tex] = standard deviation = 5 million dollars

So, probability that a randomly selected firm will earn less than 100 million dollars is given by = P(X < 100 million dollars)

P(X < 100) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{100-95}{5}[/tex] ) = P(Z < 1) = 0.8413 {using z table]

Therefore, probability that a randomly selected firm will earn less than 100 million dollars is 0.8413.