Question

suppose the mean income of firms in the industry for a year is 45 million dollars with a standard deviation of 19 million dollars. if incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn between 54 and 69 million dollars? round your answer to four decimal places.

Explanation:

Step1: Calculate the z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 45\) (mean), \(\sigma=19\) (standard deviation).
For \(x = 54\):
\(z_1=\frac{54 - 45}{19}=\frac{9}{19}\approx0.47\)
For \(x = 69\):
\(z_2=\frac{69 - 45}{19}=\frac{24}{19}\approx1.26\)

Step2: Use the standard normal distribution table

We want to find \(P(0.47<Z<1.26)\).
Since \(P(a < Z < b)=P(Z < b)-P(Z < a)\)
From the standard - normal table, \(P(Z < 0.47)=0.6808\) and \(P(Z < 1.26)=0.8962\)

Step3: Calculate the probability

\(P(0.47<Z<1.26)=P(Z < 1.26)-P(Z < 0.47)\)
\(P(0.47<Z<1.26)=0.8962 - 0.6808=0.2154\)

Answer:

\(0.2154\)