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ANSWER
[tex]71.56[/tex]EXPLANATION
Parameters given:
Mean, μ = 64
Standard deviation, σ = 9
To find the cutoff score, we want to find the score that has a corresponding z-score which represents the top 20% of the data.
To do this, first, we have to subtract 20% from 100%:
[tex]\begin{gathered} 100-20 \\ \Rightarrow80\% \end{gathered}[/tex]Now, we have to use the standard normal table to find the z score that corresponds to the closest value to 80% (0.80) on the standard normal table i.e. P(x > 80).
From the table, we see that the z-score that corresponds to 0.80 (0.79955 from the table) is 0.84.
Now, using the formula for z-score, find the cutoff score:
[tex]z=\frac{x-\mu}{\sigma}[/tex]where x = cutoff score
Solving for x, we have that:
[tex]\begin{gathered} 0.84=\frac{x-64}{9} \\ \Rightarrow0.84\cdot9=x-64 \\ \Rightarrow x-64=7.56 \\ \Rightarrow x=64+7.56 \\ x=71.56 \end{gathered}[/tex]That is the cutoff score.
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