Question

Suppose the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 11 million dollars. If incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn less than 101 million dollars? Round your answer to four decimal places, Tables Keypad Answer 2 Points If you would like to look up the value in a table, select the table you want to view, then either click the cell at the intersection of the row and column or use the arrow keys to find the appropriate cell in the table and select it using the Space key. Normal Table-to- Normal Table - toz Du

Answer 1

The mean income of firms in the industry for a year is 95 million dollars and a standard deviation of 11 million dollars.

Income for the industry are distributed normally.

We have to find the probability that a randomly selected firm will earn less than 101 million dollars.

Now, let X be the random variable denoting the income of a randomly selected firm in millions of dollars, then X follows normal with mean 95 and standard deviation of 11.

So, we can say that, Z=(X-95)/11 follows standard normal with mean 0 and standard deviation of 1.

We have to find

$=P(\frac{X-95}{11}<\frac{101-95}{11})$

$=P(Z<0.54)$

$=phi(0.54)$

Where, phi is the distribution function of the standard normal variate.

From the standard normal table, this becomes

$=0.7073$

The answer is

The approximate probability that a randomly selected firm will earn less than 101 million dollars, is 0.7073.