Question

Forgetting how to do some AP Stats work.

Scores on a standardized test have a normal distribution with a mean of 20 and a standard deviation of 6. What is the approximate interquartile range of test scores?

A. 4
B. 6
C. 8 <--
D. 10
E. 24

Answer 1

The interquartile range is the value \(k\) that gives \(\mathbb P(|Z|<k)=50\%\) where \(Z\) follows the standard normal distribution.

This is the same as
\[\begin{align*}
\mathbb P(|Z|<k)&=\mathbb P(-k<Z<k)\\[1ex]
&=\mathbb P(Z<k)-\mathbb P(Z<-k)\\[1ex]
&=75\%-25\%=50\%
\end{align*}\]This means the third quartile is \(Q_3=k\) and the first quartile is \(Q_1=-k\). Do you know how to find these?

Answer 2

@HolsterEmission How many standard deviations away from the mean is Q1 and Q3?

Answer 3

For a normal distribution, about \(0.67\) according to this: http://www.wolframalpha.com/input/?i=inversecdf(normal(0,1),+.75)

Answer 4

Ok so that's not how you do it.

Answer 5

*What's* not how you do it? The approach I outlined above works fine.
\[\mathbb P(Z<k)=0.75\implies Q_3=k\]Maybe you're confused about the fact that this is the third quartile for \(Z\), and not the distribution you're working with? To transform to the corresponding quartile for your distribution, all you need to do is use the formula \(Z=\dfrac{X-20}{6}\), or \(X=6Z+20\).

Then the quartiles for \(X\) would be \(6k+20\).

Answer 6

Sorry, the *third* quartile would be \(6k+20\). The first would be \(-6k+20\).

Then the interquartile range would be (with respect to X) \(Q_3-Q_1=(6k+20)-(-6k+20)=12k\), and \(k=0.67\), which is why (c) is the correct answer.

Answer 7

Sorry you way confuses me. I remember that I had to use InvNorm on the calculator to obtain the Zscore by plugging in .75 for area and then .25 and subtract Q1 from Q3 by solving that equation.