Given that
The mean income of firms in the industry for a year is $80 million with a standard deviation of $13 million. and we have to find the probability that a randomly selected firm will earn less than $96 million.
Explanation -
We have to find the probability that a firm will earn less than $96 million.
The mean is $80 and the standard deviation is $13.
Then, it is written as
[tex]\begin{gathered} P(x<96)=P(z<\frac{96-80}{13}) \\ \\ The\text{ formula to find the z is \lparen here z is the z value\rparen} \\ z=\frac{x-\mu}{\sigma} \\ \mu\text{ is mean and }\sigma\text{ is the standard deviation.} \\ \\ P(<96)=P(z<\frac{16}{13})=P(z<1.2) \end{gathered}[/tex]
The table to find the z value is
According to the z table, the value will be
[tex]\begin{gathered} P(x<96)=P(z<1.2)=0.8849 \\ P(x<96)=88.49\% \end{gathered}[/tex]
Hence, 88.49% of the randomly selected firms will earn less than $96.
So the probability will be 0.8849.
Final answer -
The final answer is 0.8849 or 88.49%.
