Question

at most colleges and universities, weighted - means are used to compute students grade - point averages (gpas). at one college, the grades a - f are assigned numerical values as follows: a = 4.0, c = 2.0, b+ = 3.5, d+ = 1.5, b = 3.0, d = 1.0, c+ = 2.5, f = 0.0. grade - point average is a weighted mean where the \weights\ for each grade are the number of credit hours for that class. compute the gpa for the following grades, round to the nearest hundredth. course | grade | number of credit hours (weights) intermediate algebra | b+ | 3 theater | c | 3 music appreciation | d | 3 world history | d+ | 3. the gpa for these grades is

Explanation:

Step1: Identify grade - point values

Based on the given scale: $A = 4.0$, $A-=3.7$, $B + = 3.3$, $B=3.0$, $B - = 2.7$, $C+=2.3$, $C = 2.0$, $C-=1.7$, $D+=1.3$, $D = 1.0$, $D-=0.7$, $F = 0.0$. For Intermediate Algebra with a grade of $B+$, the grade - point value $x_1=3.3$ and weight $w_1 = 3$; for Theater with a grade of $C$, the grade - point value $x_2 = 2.0$ and weight $w_2=1$; for Music Appreciation with a grade of $D$, the grade - point value $x_3 = 1.0$ and weight $w_3 = 3$; for World History with a grade of $D+$, the grade - point value $x_4=1.3$ and weight $w_4 = 1$.

Step2: Calculate the numerator of the weighted - mean formula

The formula for the weighted mean $\bar{x}=\frac{\sum_{i = 1}^{n}w_ix_i}{\sum_{i = 1}^{n}w_i}$. The numerator $\sum_{i = 1}^{4}w_ix_i=w_1x_1 + w_2x_2+w_3x_3+w_4x_4=(3\times3.3)+(1\times2.0)+(3\times1.0)+(1\times1.3)$.

• Calculate each product: $3\times3.3 = 9.9$, $1\times2.0 = 2.0$, $3\times1.0 = 3.0$, $1\times1.3 = 1.3$.
• Sum them up: $9.9+2.0 + 3.0+1.3=16.2$.

Step3: Calculate the denominator of the weighted - mean formula

The denominator $\sum_{i = 1}^{4}w_i=3 + 1+3 + 1=8$.

Step4: Calculate the GPA

The GPA (weighted mean) $\bar{x}=\frac{16.2}{8}=2.025\approx2.03$.

Answer:

$2.03$