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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 dol
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Answer #1

The mean income of firms in the industry for a year is 95 million dollars and a standard deviation of 11 million dollars.

Income for the industry are distributed normally.

We have to find the probability that a randomly selected firm will earn less than 101 million dollars.

Now, let X be the random variable denoting the income of a randomly selected firm in millions of dollars, then X follows normal with mean 95 and standard deviation of 11.

So, we can say that, Z=(X-95)/11 follows standard normal with mean 0 and standard deviation of 1.

We have to find

P(X<101)

=P(\frac{X-95}{11}<\frac{101-95}{11})

=P(Z<0.54)

=phi(0.54)

Where, phi is the distribution function of the standard normal variate.

From the standard normal table, this becomes

=0.7073

The answer is

The approximate probability that a randomly selected firm will earn less than 101 million dollars, is 0.7073.

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