Question

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

Answer 1

Step-by-step explanation:

Step 1:

We will calculate the z-score using the formula below

`\begin{gathered} z=(x-\mu)/(\sigma) \\ where. \\ \mu=70million \\ \sigma=19million \\ x=76million \end{gathered}`

By substituting the values, we will have

`\begin{gathered} z=(x-\mu)/(\sigma) \\ z=(76-70)/(19) \\ z=(6)/(19) \\ z=0.3158 \end{gathered}`

Step 2:

We will have to calculate the probabaility below

Using a p-valu calculat

`\begin{gathered} p=0.3761 \\ hence, \\ the\text{ percentage will be} \\ 0.3761*100=37.61\% \end{gathered}`

Hence,

The final answer to the nearest hundredth is

`\begin{equation*} 37.61\% \end{equation*}`