Question

Scores on a test are normally distributed with a mean of 63 and a standard deviation of 9.3. Find P81, which separates the bottom 81% from the top 19%.

Answer 1

We want to find
`x` such that

`\mathrm{Pr}(X \le x) = 0.81`

where
`X` is the random variable for test scores, and
`X\sim\mathrm{Normal}(63,9.3^2)`.

Transform
`X` to
`Z\sim\mathrm{Normal}(0,1)` using the relation

`X = \mu + \sigma Z \implies Z = \drac{X-63}{9.3}`

so that

`\mathrm{Pr}(X \le x) = \mathrm{Pr}\left((X-63)/(9.3) \le (x-63)/(9.3)\right) = \mathrm{Pr}\left(Z \le (x-63)/(9.3)\right) = 0.81`

Applying the inverse CDF of
`Z` (denoted by
`\Phi^(-1)`), we have

`(x-63)/(9.3) = \Phi^(-1)(0.81) \approx 0.8779`

Solve for
`x`.

`(x-63)/(9.3) \approx 0.8779`

`x-63 \approx 8.1644`

`x \approx 76.1644`

Then the cutoff test score for the 81% percentile is
`P_(81) \approx \boxed{76}`.