Question

The total scores on the Medical College Admission Test (MCAT) in 2013 follow a Normal distribution with mean 25.3 and standard deviation 6.5.

(a) What are the median and the fi rst and third quartiles of the MCAT scores? What is the interquartile range?
(b) Give the interval that contains the central 80% of the MCAT scores.

Answer 1

The median of the MCAT scores is equal to the mean. The first quartile and third quartile can be found using the z-score formula and the standard normal distribution table. The interquartile range is the difference between the third quartile and the first quartile. The interval that contains the central 80% of the MCAT scores can be found using the z distribution.

Step-by-step explanation:

To find the median of the MCAT scores, we first need to calculate the z-score for the median. The z-score formula is z = (x - μ) / σ, where x is the value being standardized, μ is the mean, and σ is the standard deviation. For the median, we have x = 25.3, μ = 25.3, and σ = 6.5. Plugging these values into the formula, we get z = (25.3 - 25.3) / 6.5 = 0. This means the median of the MCAT scores is equal to the mean of 25.3.

To find the first and third quartiles, we need to calculate the z-scores for these percentiles. The first quartile corresponds to the 25th percentile, and the third quartile corresponds to the 75th percentile. Using the z-score formula and the standard normal distribution table, we can find the values of x that correspond to these percentiles. The interquartile range is the difference between the third quartile and the first quartile.

For part (b), the interval that contains the central 80% of the MCAT scores can be found using the z distribution. We need to find the z-scores that correspond to the lower and upper percentiles of 10% and 90%. Using the standard normal distribution table, we can find the values of x that correspond to these z-scores. The interval is obtained by adding and subtracting the appropriate standard deviations from the mean.

Answer 2

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<Q_1)=0.25`

You have


`\mathbb P(X<Q_1)=\mathbb P\left((X-25.3)/(6.5)<(Q_1-25.3)/(6.5)\right)=\mathbb P(Z<z)=0.25`

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


`(Q_1-25.3)/(6.5)=-0.6745\implies Q_1\approx20.9`

`(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<Z<z)=0.80`

That leaves 10% on either side of this range, which means


`\underbrace{\mathbb P(-z<Z<z)}_(80\%)=\underbrace{\mathbb P(Z<z)}_(90\%)-\underbrace{\mathbb P(Z<-z)}_(10\%)`

You have


`\mathbb P(Z<z)=0.90\implies z\approx1.2816\implies -z=-1.2816`

Converting to MCAT scores,


`-1.2816=\frac{x_{\text{low}}-25.3}{6.5}\implies x_{\text{low}}\approx17.0`

`1.2816=\frac{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).