Question

percentages to
aw data\

1. the distribution of act scores is approximately normal with a mean \\( \bar{x}=21.5 \\) and a standard deviation of

\\( s = 5.4 \\).
(a) harvard university only accepts applicants in the top \\( 15 \\% \\) of all act score takers. what is the cutoff act
score?
at least a 27.1 you have to get in you want to have
a chance of getting in.
(b) the university of illinois aims to take the middle \\( 60 \\% \\) of act scores in their applications. what are the
cutoff act scores?
(c) what act scores make up the bottom \\( 10 \\% \\) of all scores?

Explanation:

Step1: Find the z - score for part (a)

We know that if we want the top \(15\%\), then the area to the left is \(A = 1-0.15=0.85\). Using a standard normal table (or a calculator with a normal - distribution function, e.g., invNorm on a TI - 84), the \(z\) - score corresponding to an area of \(0.85\) is \(z\approx1.04\).
The formula for the \(z\) - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 21.5\), \(\sigma=5.4\). Rearranging for \(x\) gives \(x=\mu + z\sigma\).
Substitute the values: \(x = 21.5+1.04\times5.4\)

$$x=21.5 + 5.616=27.116\approx27.1$$

Step2: Find the z - scores for part (b)

For the middle \(60\%\) of the data, the area in the two tails is \(1 - 0.60=0.40\), and the area in each tail is \(\frac{0.40}{2}=0.20\).
The \(z\) - score corresponding to an area of \(0.20\) (left - tail) is \(z_1\approx - 0.84\) and the \(z\) - score corresponding to an area of \(1 - 0.20 = 0.80\) (right - tail) is \(z_2\approx0.84\)
Using \(x=\mu+z\sigma\)
For \(z=-0.84\): \(x_1=21.5+( - 0.84)\times5.4=21.5-4.536 = 16.964\approx17.0\)
For \(z = 0.84\): \(x_2=21.5+0.84\times5.4=21.5 + 4.536=26.036\approx26.0\)

Step3: Find the z - score for part (c)

For the bottom \(10\%\), the area to the left is \(A = 0.10\). Using a standard normal table (or invNorm), the \(z\) - score corresponding to an area of \(0.10\) is \(z\approx - 1.28\)
Using \(x=\mu+z\sigma\), with \(\mu = 21.5\), \(\sigma = 5.4\)
\(x=21.5+( - 1.28)\times5.4=21.5-6.912 = 14.588\approx14.6\)

Answer:

(a) The cutoff ACT score is approximately \(27.1\)
(b) The cutoff ACT scores are approximately \(17.0\) and \(26.0\)
(c) The ACT scores that make up the bottom \(10\%\) are approximately \(14.6\)