C = 5(f-32) / 9 in form ac + b / c
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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.