Question

Quartiles for GPA In Example P.31 on page 728 we see that the grade point averages (GPA) for students in introductory statistics at one college are modeled with a \(\mathrm{N}(3.16,0.40)\) distribution. Find the first and third quartiles of this normal distribution. That is, find a value \(\left(\mathrm{Q}_{1}\right)\) where about \(25 \%\) of the GPAs are below it and a value \(\left(\mathrm{Q}_{3}\right)\) that is larger than about \(75 \%\) of the GPAs.

Short answer

The first quartile \( Q1 \) and the third quartile \( Q3 \) for the Normal distribution (\( \mu = 3.16 \), \( \sigma = 0.40 \)) can be calculated using the inverse cumulative distribution function for the respective percentiles 0.25 and 0.75. Use these calculations to find \( Q1 \) and \( Q3 \)

Step-by-step solution

Understand the Given Distribution

The given distribution is Normal with mean \( \mu = 3.16 \) and standard deviation \( \sigma = 0.40 \). The first quartile \( (Q1) \) and the third quartile \( (Q3) \) are required where 25% of GPAs are below \( Q1 \) and 75% of GPAs are below \( Q3 \).

Find the First Quartile (Q1)

To find the first quartile \( Q1 \), which is the cut point below which approximately 25% of observations fall, use the Normal distribution quantile function with percentile 0.25. Using a common statistical tool or calculator that supports quantile function for Normal distribution: \(Q1 = \mu + \sigma * \Phi^{-1}(0.25)\) where \( \Phi^{-1}(0.25) \) is the inverse cumulative distribution for the given percentile.

Find the Third Quartile (Q3)

Similarly, to find the third Quartile \( Q3 \), which is the cut point below which approximately 75% of observations fall, use the Normal distribution quantile function with percentile 0.75. So, \(Q3 = \mu + \sigma * \Phi^{-1}(0.75)\) where \( \Phi^{-1}(0.75) \) is the inverse cumulative distribution for the given percentile.