Last year, the revenue for financial services companies had a mean of 70 million dollars with a standard deviation of 19 Find the percentage of companies with revenue greater than 76 million dollarsAssume that the distribution is normalRound your answer to the nearest hundredth

Respuesta :

Explanation:

Step 1:

We will calculate the z-score using the formula below

[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ where. \\ \mu=70million \\ \sigma=19million \\ x=76million \end{gathered}[/tex]

By substituting the values, we will have

[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ z=\frac{76-70}{19} \\ z=\frac{6}{19} \\ z=0.3158 \end{gathered}[/tex]

Step 2:

We will have to calculate the probabaility below

Using a p-valu calculat

[tex]\begin{gathered} p=0.3761 \\ hence, \\ the\text{ percentage will be} \\ 0.3761\times100=37.61\% \end{gathered}[/tex]

Hence,

The final answer to the nearest hundredth is

[tex]\begin{equation*} 37.61\% \end{equation*}[/tex]