Question

5#The distribution of scores on a standardized aptitude test is approximately normal with a mean of 480 and a standard deviation of 105. What is the minimum score needed to be in the top 20% on this test? Carry your intermediate computations to at least four decimal places, and round your answer to the nearest integer.

Answer 1

We want to find the value that makes

$P(X\ge x)=0.2$

To find it we must look at the standard normal table, using the complementary cumulative table we find that

$P(Z\ge z)=0.20045\Leftrightarrow z=0.84$

Then, using the z-score we can find the minimum score needed, remember that

$z=\frac{x-\mu}{\sigma}$

Where

σ = standard deviation

μ = mean

And in our example, x = minimum score needed, therefore

$\begin{gathered} 0.84=\frac{x-480}{105} \\ \\ x=0.84\cdot105+480 \\ \\ x=568.2 \end{gathered}$

Rounding to the nearest integer the minimum score needed is 568, if you get 568 you are at the top 20.1%.