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Suppose the mean income of firms in the industry for a year is 80 million dollars with a standard deviation of 11 million dollars. If incomes for theindustry are distributed normally, what is the probability that a randomly selected firm will earn less than 103 million dollars? Round your answer tofour decimal places.

Respuesta :

Mean is 80 million

Standar deviation is 11 million

Probabiliy of less tha 103 million

First we normalize the values using the fomula:

Z = (x - u)/s

x = 103

u = 80

s = 11

Z = (103-80)/11 = 23/11 = 2.09

Using the standar normal ditribution tables, this value represents: 0.01831

So the probability is 1 - 0.01831 - 0.9817

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