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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.



Answer :

The probability that a randomly selected firm will earn less than 101 million dollars will be;

⇒ 0.5454

What is Standard deviation?

Standard deviation is the measure of dispersed the data is in relation to the mean. Low standard deviation means data are clustered around the mean, and high standard deviation indicates data are more spread out.

Given that;

The mean income of firms in the industry = 95 million dollars.

And, Standard deviation = 11 million dollars.

Now, The z - score probability distribution is given as;

Z = X - μ / σ

Where,  μ is the mean income of firms in the industry.

And, σ is standard deviation.

So, The probability that a randomly selected firm will earn less than 101 million dollars will be;

P (Z < 101) = P ( X - μ / σ < 101 - 95 / 11 )

P (Z < 101) = 0.5454

Therefore, The probability that a randomly selected firm will earn less than 101 million dollars will be;

⇒ 0.5454

Learn more about the z - score visit:

https://brainly.com/question/25638875

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