Word Questions on Mean of Raw Data helps you all to find the mean (or average) of an ungrouped data set where the data is not presented in intervals. Moreover, you can learn more knowledge about the concept by practicing the example problems based on Mean of ungrouped data.

Here, we have explained various styles of word problems on Mean of arrayed data or ungrouped data for your knowledge. Take a look at them and ace up your subject preparation by answering the Mean Deviation for ungrouped data examples and cross-check the solutions here itself.

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### Mean of Ungrouped Data Example Problems PDF and Solutions

**Example 1:
**The monthly salary (in $) of 8 employees in a company are 6000, 8000, 4000, 9000, 5000, 3000, 8000, 6500. Find the mean of ungrouped data?

**Solution:**

Given salary (in $) for 8 employees are 6000, 8000, 4000, 9000, 5000, 3000, 8000, 6500

Mean of ungrouped data = Sum of the observations / Total number of observations

= 6000+8000+4000+9000+5000+3000+8000+6500 / 8

= 49500 / 8

= 6187.5

**Example 2: **

The mean age of ten girls is 25 years. If the ages of nine of them be 10 years, 12 years, 13 years, 15 years and 25 years then find the age of the tenth girl.

**Solution:**

Let the age of the tenth girl be x years

Now, find the mean age of 10 girls = 10 years +12 years +13 years +15 years + 25 years + x years / 10

⟹ 25 = 75 years + x years / 10 (mean age of 10 girls is 25 years from the question)

⟹ 250 = 75 + x

⟹ x = 250-75

⟹ x = 175

Hence the age of the tenth girl is 175 years.

**Example 3:
**Find the mean of the first four whole numbers.

**Solution:**

The first four whole numbers are 0, 1, 2, 3.

Hence, the mean = x1+x2+x3+x4 / 4

= 0+1+2+3/4

= 6/4

= 3/2

= 1.5

**Example 4: **

The mean of a sample of 5 numbers is 4. An extra value of 2.5 is added to the sample then find the new mean?

**Solution:**

Given that Total of original numbers =5×4=20

New total =20+2.5=22.5

The new mean of a sample = 22.5/6

= 3.75

Hence, the new mean of 6 numbers is 3.75.

**Example 5: **

The following table gives the points of each player scored in four games:

Player |
Game 1 |
Game 2 |
Game 3 |
Game 4 |

A
B C |
14
0 8 |
16
8 11 |
10
6 Did not play |
10
4 13 |

(i) Find the mean to determine C’s average number of points scored per game.

(ii) Who is the worst performer?

**Solution:**

(i) Mean Score of C = 8+11+13 / 3

= 32/3

= 10.6

(ii) To calculate who is the worst performer, we have to find the mean score of each player:

Mean score of A = (14 + 16 + 10 + 10) / 4 = 12.5

Mean score of B = (0 + 8 + 6 + 4) / 4 = 18/4 = 4.5

Mean Score of C = 8+11+13 / 3 = 32/3 = 10.6

Therefore, B is the worst performer.

**Example 6:
**A competitor scored 80%, 90%, 75%, and 60% in four subjects in an entrance test, Calculate the mean percentage of the scores achieved by his/her.

**Solution:**

Given observations in percentage are x1=80, x2=90, x3=75, x4=60

Hence, their mean A = x1+x2+x3+x4 / 4

= 80+90+74+60 / 4

= 304 / 4

= 76

Hence, the mean percentage of scores achieved by the competitor was 76%.