Question

Help Please!
A survey was taken of students in math classes to find out how many hours per day students spend on social media. The survey results for the first-, second-, and third-period classes are as follows:

First period: 2, 4, 3, 1, 0, 2, 1, 3, 1, 4, 9, 2, 4, 3, 0

Second period: 3, 2, 3, 1, 3, 4, 2, 4, 3, 1, 0, 2, 3, 1, 2

Third period: 4, 5, 3, 4, 2, 3, 4, 1, 8, 2, 3, 1, 0, 2, 1, 3

Which is the best measure of center for first period and why?
a. Mean, because there are no outliers that affect the center
b. Median, because there is 1 outlier that affects the center
c. Interquartile range, because there is 1 outlier that affects the center
d. Standard deviation, because there are no outliers that affect the center

Answer 1

Concept

This problem is related to Mean, median, and standard deviation which are important tools in the statistics. These measures of the center are used to approximate and understand a “middle value” or “average” of a given data set.

Given

We have given that a survey was taken of students in math classes to find out how many hours per day students spend on social media. The survey results for the first-, second-, and third-period classes are as follows:

First period: 2, 4, 3, 1, 0, 2, 1, 3, 1, 4, 9, 2, 4, 3, 0

Second period: 3, 2, 3, 1, 3, 4, 2, 4, 3, 1, 0, 2, 3, 1, 2

Third period: 4, 5, 3, 4, 2, 3, 4, 1, 8, 2, 3, 1, 0, 2, 1, 3

To find

We have to find Which is the best measure of center for the first period and why

a. Mean, because there are no outliers that affect the center

b. Median, because there is 1 outlier that affects the center

c. Interquartile range, because there is 1 outlier that affects the center

d. Standard deviation, because there are no outliers that affect the center.

Solution

Answer is option (a)

Here,

First  period:  $2, 4, 3, 1, 0, 2, 1, 3, 1, 4, 9, 2, 4, 3, 0$

Firstly, arrange data in increasing order as,

$0, 0, 1, 1, 1, 2,2,2,3,3,3, 4, 4,4,9$

Here, for the mean Add all the values and divide by the number of values.

$\mu=\frac{0+0+1+1+1+2+2+2+3+3+3+4+4+4+9}{15}$

$\mu=\frac{39}{15} =2.6$

And then identify the middle value, here we have seven values to the right of 2 and seven values to the left of 2. The middle value is $2$,

So, the median is $2$

As a result, the Mean is the best measure of the center for the first period, because there are no outliers that affect the center.

#SPJ3

Answer 2

Answer:

Step-by-step explanation:

i refer you to look at another page. There is the same question asked, but with the right answer.

https://study.com/academy/answer/a-survey-was-taken-of-students-in-math-classes-to-find-out-how-many-hours-per-day-students-spend-on-social-media-the-survey-results-for-the-first-second-and-third-period-classes-are-as-follows-first-period-2-4-3-1-0-2-1-3-1-4-9-2-4-3.html