Question

The Arc Electronic Company had an income of 9090 million dollars last year. Suppose the mean income of firms in the same industry as Arc for a year is 5555 million dollars with a standard deviation of 1717 million dollars. If incomes for this industry are distributed normally, what is the probability that a randomly selected firm will earn more than Arc did last year? Round your answer to four decimal places.

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Answer 1

This is a normal distribution question with

P(x > 9090.0)=?

The z-score at x = 9090.0 is,

z = 2.0588

This implies that

P(x > 9090.0) = P(z > 2.0588) = 1 - 0.9802432991597723

PS: you have to refer z score table to find the final probabilities.

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