HOW TO SOLVE MEAN MEDIAN AND MODE (UNGROUPED)

THE MEAN MEDIAN AND MODE (UNGROUPED)

In measures of central tendency, there are two types of grouping, "grouped data" and "ungrouped data". In grouped data, it is said as a recommendation that the data must be 30 and above, and 29 and below for ungrouped data.

Mean is the most commonly used among the three (mean, median, mode) since this is by getting the average. It is done by summing up the total data and dividing them by how many those data are. Say, the total weight of five students is the combined weight of five students themselves divided by five.

Having a median is a little different. The procedures are arranged by data in ascending or descending order and getting what is in the middle speaks for the median. Say, 3,6,7,9,10, and 7 are the median as it was in the center of ascending data.

Getting the mode plays differently. Frequency from the word "frequent" hints here. In, 3,2,2,5,6, only (2) was the mode since 2 appears more frequently than others. In 3,4,4,5,6,7,6,8,9, It is 4, and 6 appears more frequently, so 4 and 2 are the modes and it is named "bi-modal". The data is bi-modal.

Illustrative Examples:

Find the Mean, Median, and Mode of the data given:

Data: 3,5,7,2,4,5,9,1,6,3

Solutions:

$1.)Mean=\frac{3+5+7+2+4+5+9+1+6+3}{10}=\frac{45}{10}={\color{Red} 4.5}$

$2.)Median=1,2,3,3,{\color{Red} 4},{\color{Red} 5},5,6,7,9=\frac{4+5}{2}=\frac{9}{2}={\color{Red} 4.5}$

Notice that the middle terms were 4 and 5 as the data was arranged in ascending order. To have the median, getting the average of the two middles shows the result of 4.5. This is so when the data is even.

$3.) Mode={\color{Red} 3}\&{\color{Red} 5}$ , So 3 and 5 have more appearance than the others. It is bi-modal.

More Illustrations:

Given the data, 12,34,15, 17,10, Find the Mean, Median, and Mode.

$1.) Mean=\frac{12+34+15+17+10}{5}=\frac{88}{5}={\color{Red} 17.6}$

$2.) Median=10,12,{\color{Red} 15},17,34$ , Our median is 15.

$3.) Mode=(No\ Mode)$ This time, no more data appears than the others. There's no mode to this data.

Problem:

There were 8 students with the following weight in kilograms. The data is shown below.

$\begin{bmatrix} Student\ Name & Weight(kg)\\ Greg&56 \\ John&52 \\ Douglas&45 \\ Johanna&60 \\ Paul& 67\\ Mariel&40 \\ Anna&45 \\ Jacky& 53 \end{bmatrix}$

Find the Mean, Median, and Mode of the data given in the table above.

For Mean, add the data and divide it into how many they are in the data set.

$1.) Mean= \frac{56+52+45+60+67+40+45+53}{8}=\frac{418}{8}={\color{Red} 52.25}$

For Median, arrange the data in ascending or descending order. Here, we prefer to arrange it in ascending order.

$2.) Median=40,45,45,{\color{Red} 52},{\color{Red} 53},56,60,67=\frac{52+53}{2}=\frac{105}{2}={\color{Red} 52.5}$

For Mode, find the most data appearing data in a data set.

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Challenge yourself:

What is the Mean, Median, and Mode if the data are: 3,6,7,4,6? Write your answer on the "Post a comment" box below this page.

1.) Mean =?

2.) Mode =?

3.) Median =?

STATISTICS

+ comments + 5 comments

Mickaela Macaraeg

9:43 PM MST

Mean:
3,6,7,4,6
= 26÷5
= 5.2
Median:
=7
Mode:
=6
Range:
3,4,6,6,7
7-3
=4

Mickaela P. Macaraeg

9:54 PM MST

FINAL ANSWER

1. Mean = (3 + 6 + 7 + 4 + 6) / 5 = 26 / 5 = 5.2

2. Mode = 6

3. Median = 6

Faith Casañares

10:16 PM MST

Mean = 3+6+7+4+6 = 26 ÷ 5 = 5.2
Median = 3, 4, 6, 6, 7 = 6
Mode = 6

Rogelio Andrei G. Amanaia

5:40 AM MST

1. Mean = 3 + 6 + 7 + 4 + 6 = 26
Divide by the number of values
= 26 ÷ 5 = 5.2
So Mean is equal to 5.2

2. Median = 3, 4, 6, 6, 7 ( arrange to ascending order)
Median = 6 (since the order is an odd count the median is in the middle)

3. Mode = 3, 4, 6, 6, 7
The 6 appears twice while the others appears only once.
Mode = 6

Anonymous

4:55 AM MST

Angelica Jasmine Palaran
Mean=3+6+7+4+6=26÷5=5.2
Median=3,4,6,6,7=6
Mode=6

DMG