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Supposed to mean income of firms in the industry for year is $85 million with a standard deviation of $9 million. If incomes for the industry are distributed normally what is the probability that a randomly selected firm will earn less than $105 million? Round answer to four decimal places

Supposed to mean income of firms in the industry for year is $85 million with a standard-example-1

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5 votes
Given
Z=(x-\mu)/(\sigma)
\begin{gathered} x=105miliion \\ \mu=85\text{ million} \\ \sigma=9\text{ million} \end{gathered}
\begin{gathered} Z=(105-85)/(9) \\ \\ Z=2.22222 \end{gathered}

Now

The final answer
0.9869\text{ \lparen4 d. p\rparen}

Supposed to mean income of firms in the industry for year is $85 million with a standard-example-1
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