Question

To quality for a police academy, applicants are given a lest of physical fitness. The scores are normally distributed with a mean of 64 and a standard deviation of 9. If only the top 20% of the applicants are selected, find the cutoff score.

Answer 1

`71.56`

Step-by-step 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%:

`\begin{gathered} 100-20 \\ \Rightarrow80\% \end{gathered}`

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:

`z=(x-\mu)/(\sigma)`

where x = cutoff score

Solving for x, we have that:

`\begin{gathered} 0.84=(x-64)/(9) \\ \Rightarrow0.84\cdot9=x-64 \\ \Rightarrow x-64=7.56 \\ \Rightarrow x=64+7.56 \\ x=71.56 \end{gathered}`

That is the cutoff score.