Question

the undergraduate grade point averages (ugpa) of students taking an admissions test in a recent year can be approximated by a normal distribution, as shown in the figure. (a) what is the minimum ugpa that would still place a student in the top 15% of ugpas? (b) between what two values does the middle 50% of the ugpas lie? (a) the minimum ugpa that would still place a student in the top 15% of ugpas is (round to two decimal places as needed.)

Explanation:

Step1: Find the z - score for the top 15%

The area to the left of the z - score we want is \(1 - 0.15=0.85\). Looking up in the standard normal distribution table (z - table), the z - score \(z\) corresponding to an area of 0.85 is approximately \(z = 1.04\).

Step2: Use the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 3.34\) is the mean and \(\sigma=0.16\) is the standard deviation. We want to solve for \(x\). Rearranging the formula gives \(x=\mu + z\sigma\).
Substitute \(\mu = 3.34\), \(z = 1.04\) and \(\sigma = 0.16\) into the formula: \(x=3.34+1.04\times0.16\).

Step3: Calculate the value of \(x\)

First, calculate \(1.04\times0.16 = 0.1664\). Then \(x=3.34 + 0.1664=3.5064\approx3.51\).

For part (b):

Step1: Find the z - scores for the middle 50%

The middle 50% means that the area in the two tails is \(1 - 0.50 = 0.50\), so the area in each tail is \(\frac{0.50}{2}=0.25\).
The z - score corresponding to an area of \(0.25\) in the left - hand tail is \(z_1\approx - 0.67\) and the z - score corresponding to an area of \(0.25 + 0.50=0.75\) (right - hand side of the middle 50%) is \(z_2\approx0.67\).

Step2: Use the z - score formula for the lower value

Using \(z_1=\frac{x_1-\mu}{\sigma}\), and solving for \(x_1\) gives \(x_1=\mu+z_1\sigma\). Substitute \(\mu = 3.34\), \(z_1=-0.67\) and \(\sigma = 0.16\): \(x_1=3.34+( - 0.67)\times0.16=3.34 - 0.1072 = 3.2328\approx3.23\).

Step3: Use the z - score formula for the upper value

Using \(z_2=\frac{x_2-\mu}{\sigma}\), and solving for \(x_2\) gives \(x_2=\mu+z_2\sigma\). Substitute \(\mu = 3.34\), \(z_2 = 0.67\) and \(\sigma = 0.16\): \(x_2=3.34+0.67\times0.16=3.34 + 0.1072=3.4472\approx3.45\).

Answer:

(a) 3.51
(b) 3.23 and 3.45