The ages of the students of groups A and B are displayed in the histograms.
For group A
We can determine the number of students per age by looking at the bars of the histogram
2 are 15 years old
5 are 16 years old
6 are 17 years old
5 are 18 years old
2 are 19 years old
The total students for group A is
[tex]\begin{gathered} n_A=2+5+6+5+2 \\ n_A=20 \end{gathered}[/tex]To calculate the average age on group a you have to use the following formula
[tex]X^{\text{bar}}=\frac{\Sigma x_if_i}{n}[/tex]Σxifi indicates the sum of each value of age multiplied by its observed frequency
n is the total number of students of the group
For group A the average value is
[tex]\begin{gathered} X^{\text{bar}}_A=\frac{(15\cdot2)+(16\cdot5)+(17\cdot6)+(18\cdot5)+(19\cdot2)}{20} \\ X^{\text{bar}}_A=\frac{340}{20} \\ X^{\text{bar}}_A=17 \end{gathered}[/tex]The average year of group A is 17 years old.
To determine the Median of the group, you have to calculate its position first.
[tex]\begin{gathered} \text{PosMe}=\frac{n}{2} \\ \text{PosMe}=\frac{20}{2} \\ \text{PosMe}=10 \end{gathered}[/tex]The Median is in the tenth position. To determine the age it corresponds you have to look at the accumulated observed frequencies:
F(15)=2
F(16)=2+5=7
F(17)=7+6=13→ The 10nth observation corresponds to a 17 year old student
F(18)=13+5=18
F(19)=18+2=20
The median of group A is 17 years old.
For group B
As before we can determine the number of students per age by looking at the bars of the histogram
2 are 15 years old
3 are 16 years old
4 are 17 years old
5 are 18 years old
6 are 19 years old
The total number of students for group B is
[tex]\begin{gathered} n_B=2+3+4+5+6 \\ n_B=20 \end{gathered}[/tex]The average age of group B can be calculated as
[tex]\begin{gathered} X^{\text{bar}}_B=\frac{\Sigma x_if_i}{n} \\ X^{\text{bar}}_B=\frac{(2\cdot15)+(3\cdot16)+(4\cdot17)+(5\cdot18)+(6\cdot19)}{20} \\ X^{\text{bar}}_B=\frac{350}{20} \\ X^{\text{bar}}_B=17.5 \end{gathered}[/tex]The average age for group B is 17.5 years old
Same as before, to determine the median you have to calculate its position in the sample and then locate it:
[tex]\begin{gathered} \text{PosMe}=\frac{n}{2} \\ \text{PosMe}=\frac{20}{2} \\ \text{PosMe}=10 \end{gathered}[/tex]The median is in the 10nth position, to determine where the 10nth student is located you have to take a look at the accumulated frequencies:
F(15)=2
F(16)=2+3=5
F(17)=5+4=9
F(18)=9+5=14 →The 10nth observation corresponds to a 18 year old student
F(19)=14+6=20
The median of group B is 18 years old
So
[tex]\begin{gathered} X^{\text{bar}}_A=17 \\ X^{\text{bar}}_B=17.5_{} \\ Me_A=17 \\ Me_B=18_{} \end{gathered}[/tex]The mean and median of group B are greater than the mean and median from group B. The correct choice is the first one.
