Question

Six ninth-grade students and six 12th-grade students were asked: How many movies have you seen this month? Here are their responses.
Ninth-grade students: 5,1, 2, 5, 3,8
12th-grade students: 4,2,0, 2, 3, 1

a. By hand, calculate the mean, variance, and standard deviation of each of these data sets. Which is more spread out, the ninth-grade or 12th-grade data set?
b. Make a graph of both data sets. Which of these data sets appears more spread out? Does your answer agree with your conclusion in part (a)?

Answer 1

Answer:

Following are the solution to the given point:

Step-by-step explanation:

Grade of students:

Grade of student:

In point a:

Grade of $9^{th}$ students:

$\to n= 6\\\\\to Mean = \frac{ \sum x_i}{n}$

               $= \frac{ 5+1+2+5+2+8}{6}\\\\ = \frac{24}{6}\\\\ = 4$

$\to Variance\\\\\sigma^2 = \frac{\sum(x_i -\bar{x})^2}{n-1}$

    $=\frac{1}{5}(10)\\\\=2$

$\to Standard \ deviation ( \sigma) :\\\\= \sqrt{\sigma^2}\\\\= \sqrt{2}\\\\= 1.141$

Since the Grade of $9^{th}$ student's variance is higher than $12^{th}$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.