In a normal distribution, the median is the same as the mean (25.3). The first quartile is the value of \(Q_1\) such that
\(\mathbb P(X
You have
\(\mathbb P(X
For the standard normal distribution, the first quartile is about \(z\approx-0.6745\), and by symmetry the third quartile would be \(z\approx0.6745\). In terms of the MCAT score distribution, these values are
\(\dfrac{Q_1-25.3}{6.5}=-0.6745\implies Q_1\approx20.9\)
\(\dfrac{Q_3-25.3}{6.5}=0.6745\implies Q_3\approx29.7\)
The interquartile range (IQR) is just the difference between the two quartiles, so the IQR is about 8.8.
The central 80% of the scores have z-scores \(\pm z\) such that
\(\mathbb P(-z
That leaves 10% on either side of this range, which means
\(\underbrace{\mathbb P(-z
You have
\(\mathbb P(Z
Converting to MCAT scores,
\(-1.2816=\dfrac{x_{\text{low}}-25.3}{6.5}\implies x_{\text{low}}\approx17.0\)
\(1.2816=\dfrac{x_{\text{high}}-25.3}{6.5}\implies x_{\text{high}}\approx33.6\)
So the interval that contains the central 80% is \((17.0,33.6)\) (give or take).
