Suppose the mean income of firms in the industry for a year is 25 million dollars with a standard deviation of 9 million dollars. If incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn less than 34 million dollars? Round your answer to four decimal places.

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ChiKesselman ChiKesselman · Answer 1 · 2019-10-24T07:03:52+02:00

Answer:

About 84.13% of the randomly selected firm will earn less than 34 million dollars.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 25 million dollars

Standard Deviation, σ = 9 million dollars

We are given that the distribution of income for industry is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

a) P(income less than 34 million dollars)

P(x < 34)

[tex]P( x < 34) = P( z < \displaystyle\frac{34 - 25}{9}) = P(z < 1)[/tex]

Calculation the value from standard normal z table, we have,

[tex]P(x < 34) = 0.8413 = 84.13\%[/tex]

Hence, about 84.13% of the randomly selected firm will earn less than 34 million dollars.