Question

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

Answer 1

Answer:

Probability that a randomly selected firm will earn less than 100 million dollars is 0.8413.

Step-by-step explanation:

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($\mu=95,\sigma^{2}=5^{2}$)

Now, the z score probability distribution is given by;

         Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = mean income of firms in the industry = 95 million dollars

            $\sigma$ = 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( $\frac{X-\mu}{\sigma}$ < $\frac{100-95}{5}$ ) = 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.