Answer:
Following are the solution to the given point:
Step-by-step explanation:
Grade of students:
Grade of student:
In point a:
Grade of [tex]9^{th}[/tex] students:
[tex]\to n= 6\\\\\to Mean = \frac{ \sum x_i}{n} \\\\[/tex]
[tex]= \frac{ 5+1+2+5+2+8}{6}\\\\ = \frac{24}{6}\\\\ = 4[/tex]
[tex]\to Variance\\\\\sigma^2 = \frac{\sum(x_i -\bar{x})^2}{n-1}[/tex]
[tex]=\frac{1}{5}(10)\\\\=2[/tex]
[tex]\to Standard \ deviation ( \sigma) :\\\\= \sqrt{\sigma^2}\\\\= \sqrt{2}\\\\= 1.141[/tex]
Since the Grade of [tex]9^{th}[/tex] student's variance is higher than [tex]12^{th}[/tex]grade,
In point b:
Please find the attached file.
The plot above shows that the student of the ninth grade is spread more widely, as the student from 0-8 to 12 is from 0-4.
