In a normal distribution, the median is the same as the mean (25.3). The first quartile is the value of [tex]Q_1[/tex] such that
[tex]\mathbb P(X<Q_1)=0.25[/tex]
You have
[tex]\mathbb P(X<Q_1)=\mathbb P\left(\dfrac{X-25.3}{6.5}<\dfrac{Q_1-25.3}{6.5}\right)=\mathbb P(Z<z)=0.25[/tex]
For the standard normal distribution, the first quartile is about [tex]z\approx-0.6745[/tex], and by symmetry the third quartile would be [tex]z\approx0.6745[/tex]. In terms of the MCAT score distribution, these values are
[tex]\dfrac{Q_1-25.3}{6.5}=-0.6745\implies Q_1\approx20.9[/tex]
[tex]\dfrac{Q_3-25.3}{6.5}=0.6745\implies Q_3\approx29.7[/tex]
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 [tex]\pm z[/tex] such that
[tex]\mathbb P(-z<Z<z)=0.80[/tex]
That leaves 10% on either side of this range, which means
[tex]\underbrace{\mathbb P(-z<Z<z)}_{80\%}=\underbrace{\mathbb P(Z<z)}_{90\%}-\underbrace{\mathbb P(Z<-z)}_{10\%}[/tex]
You have
[tex]\mathbb P(Z<z)=0.90\implies z\approx1.2816\implies -z=-1.2816[/tex]
Converting to MCAT scores,
[tex]-1.2816=\dfrac{x_{\text{low}}-25.3}{6.5}\implies x_{\text{low}}\approx17.0[/tex]
[tex]1.2816=\dfrac{x_{\text{high}}-25.3}{6.5}\implies x_{\text{high}}\approx33.6[/tex]
So the interval that contains the central 80% is [tex](17.0,33.6)[/tex] (give or take).
