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

Following are the solution to the given point:

Explanation:

Grade of students:

Grade of student:

In point a:

Grade of
`9^(th)` students:

`\to n= 6\\\\\to Mean = ( \sum x_i)/(n) \\\\`

`= ( 5+1+2+5+2+8)/(6)\\\\ = (24)/(6)\\\\ = 4`

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

`=(1)/(5)(10)\\\\=2`

`\to Standard \ deviation ( \sigma) :\\\\= √(\sigma^2)\\\\= √(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.