Question

suppose the mean income of firms in the industry for a year is 65 million dollars with a standard deviation of 1 million dollars. if incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn less than 66 million dollars? round your answer to four decimal places.

Answer 1

industry are distributed normally, then the probability that a randomly selected firm will earn less than 66 million dollars is 0.06712

You have to calculate the probability of a single movie costing between 69 and  66 million dollars, symbolically:

P(66≤X≤69)= (X≤69)-P(X≤66)

You have to use the standard normal distribution to calculate this probability, so first you have to calculate the Z values that correspond to each value of X using: Z= (X-μ)/σ  ~ N(0;1)

Z₁= (69-65)/18= 0.22

Z₂=(66-65)/18= 0.05

Now you look for the corresponding probability values using the standard normal table

P(Z≤0.22)= 0.58706

P(Z≤0.05)= 0.51994

P(66≤X≤69)= (X≤69)-P(X≤66)

P(Z≤0.22)-P(Z≤0.05)= 0.58706 - 0.51994= 0.06712

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