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Answer:
Probability that a randomly selected firm will earn less than 100 million dollars is 0.8413.
Step-by-step explanation:
We are given that the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 5 million dollars. Also, incomes for the industry are distributed normally.
<em>Let X = incomes for the industry</em>
So, X ~ N()
Now, the z score probability distribution is given by;
Z = ~ N(0,1)
where, = mean income of firms in the industry = 95 million dollars
= standard deviation = 5 million dollars
So, probability that a randomly selected firm will earn less than 100 million dollars is given by = P(X < 100 million dollars)
P(X < 100) = P( <
) = P(Z < 1) = 0.8413 {using z table]
Therefore, probability that a randomly selected firm will earn less than 100 million dollars is 0.8413.
Remember that the radicand (the area under the root sign) must be positive or zero for a radical with an even index (like the square root or fourth root, for example). This is because two numbers squared or to the fourth power, etc. cannot be negative, so there are no real solutions when the radicand is negative. We must restrict the domain of the square-root function.
If the domain has already been restricted to
, we can work backwards to add 11 to both sides. We see that 
must be under the radicand, so the answer is A.
If the domain has already been restricted to