Question 1:
Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14
By using the empirical relation also find the mean.
Arranging the data in ascending order such that same numbers are put together, we get:
12,12,13,13, 14,14,14, 16, 19
Here, n = 9.
∴ Median = Value of observation = Value of the 5th observation = 14.
Here, 14 occurs the maximum number of times, i.e., three times. Therefore, 14 is the mode of the data.
Now,
Mode = 3 Median - 2 Mean
⇒ 14 = 3 x 14 - 2 Mean
⇒2 Mean = 42 - 14 = 28
⇒ Mean = 28 ÷ 2 = 14.
Find the median and mode of the data: 35, 32, 35, 42, 38, 32, 34
Arranging the data in ascending order such that same numbers are put together, we get:
32, 32, 34,35,35, 38,42.
Here, n = 7
∴ Median = Value of observation = Value of the 4th observation = 35.
Here, 32 and 35, both occur twice. Therefore, 32 and 35 are the two modes.
Find the mode of the data: 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4
Arranging the data in ascending order such that same values are put together, we get:
0, 2, 2, 2, 3, 3,3,4,4,4,5,5,6.
Here, 2,3 and 4 occur three times each. Therefore, 2 ,3 and 4 are the three modes.
Alternate Solution
Arranging the data in the form of a frequency table, we have:
Values | Tally Bars | Frequency |
0 | ∣ | 1 |
2 | ∣∣∣ | 3 |
3 | ∣∣∣ | 3 |
4 | ∣∣∣ | 3 |
5 | ∣∣ | 2 |
6 | ∣ | 1 |
Total | 13 |
Clearly, the values 2,3 and 4 occur the maximum number of times, i.e., three times.
Hence, the mode is 2,3 and 4.
The runs scored in a cricket match by 11 players are as follows:
6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 10
Find the mean, mode and median of this data.
Answer 4:
Arranging the data in ascending order such that same values are put together, we get:
6,8,10,10,10,15,15,50,80,100,120.
Here, n = 11
∴ Median = Value of observation = Value of the 6th observation = 15.
Here, 10 occurs three times. Therefore, 10 is the mode of the given data.
Now,
Mode = 3 Median - 2 Mean
⇒ 10 = 3 x 15 - 2 Mean
⇒2 Mean = 45 - 10 = 35
⇒ Mean = 35 ÷ 2 = 17.5.
Find the mode of the following data:
12, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14
Arranging the data in ascending order such that same values are put together, we get:
10,12,12,14,14,14,14,14,14, 16, 16, 18.
Here, clearly, 14 occurs the most number of times.
Therefore, 14 is the mode of the given data.
Alternate solution:
Arranging the data in the form of a frequency table, we get:
Values | Tally Bars | Frequency |
10 | ∣ | 1 |
12 | ∣∣ | 2 |
14 | 6 | |
16 | ∣∣ | 2 |
18 | ∣ | 1 |
Total | 12 |
Clearly, 14 has maximum frequency. So, the mode of the given data is 14.
Heights of 25 children (in cm) in a school are as given below:
168, 165, 163, 160, 163, 161, 162, 164, 163, 162, 164, 163, 160, 163, 163, 165, 163, 162, 163, 164, 163, 160, 165, 163, 162
What is the mode of heights?
Also, find the mean and median.
Arranging the data in tabular form, we get:
Height of Children (cm) | Tally Bars | Frequency |
160 | ∣∣∣ | 3 |
161 | ∣ | 1 |
162 | ∣∣∣∣ | 4 |
163 | 10 | |
164 | ∣∣∣ | 3 |
165 | ∣∣∣ | 3 |
168 | ∣ | 1 |
Total | 25 |
Here, n = 25
∴ Median = Value of observation = Value of the 13th observation = 163 cm.
Here, clearly, 163 cm occurs the most number of times. Therefore, the mode of the given data is 163 cm.
Now,
Mode = 3 Median - 2 Mean
⇒ 163 = 3 x 163 - 2 Mean
⇒2 Mean = 326
⇒ Mean = 326 ÷ 2 = 163 cm.
The scores in mathematics test (out of 25) of 15 students are as follows:
19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20
Find the mode and median of this data. Are they same?
Arranging the data in ascending order such that same values are put together, we get:
5,9,10,12,15,16, 19, 20, 20, 20, 20, 23, 24, 25, 25.
Here, n = 15
∴ Median = Value of observation = Value of the 8th observation = 20.
Here, clearly, 20 occurs the most number of times, i.e., 4 times. Therefore, the mode of the given data is 20.
Yes, the median and mode of the given data are the same.
Calculate the mean and median for the folllowing data:
Marks | : | 10 | 11 | 12 | 13 | 14 | 16 | 19 | 20 |
Number of students | : | 3 | 5 | 4 | 5 | 2 | 3 | 2 | 1 |
Using empirical formula, find its mode.
Calculation of Mean
Marks (xi) | 10 | 11 | 12 | 13 | 14 | 16 | 19 | 20 | Total |
Number of Students (fi) | 3 | 5 | 4 | 5 | 2 | 3 | 2 | 1 | |
fixi | 30 | 55 | 48 | 65 | 28 | 48 | 38 | 20 |
Mean =
Here, n = 25, which is an odd number. Therefore,
Median = Value of observation = the 13th observation = 13.
Now,
Mode = 3 Median - 2 Mean
⇒Mode = 3 x 13 - 2 x (13.28)
⇒Mode = 39 - 26.56
⇒Mode = 12.44.
The following table shows the weights of 12 persons.
Weight (in kg): | 48 | 50 | 52 | 54 | 58 |
Number of persons: | 4 | 3 | 2 | 2 | 1 |
Find the median and mean weights. Using empirical relation, calculate its mode.
Calculation of Mean
Weight (xi) | 48 | 50 | 52 | 54 | 58 | Total |
Number of Persons (fi) | 4 | 3 | 2 | 2 | 1 | |
fixi | 192 | 150 | 104 | 108 | 58 |
Here, n = 12
Now,
Mode = 3 Median - 2 Mean
⇒ Mode = 3 x 50 - 2 x 51
⇒Mode = 150 - 102
⇒ Mode = 48 kg.
Thus, Mean = 51 kg, Median = 50 kg and Mode = 48 kg.
1. What are the central values in data handling? |
2. How is the mean calculated in data handling? |
3. What is the median in data handling? |
4. How is the mode determined in data handling? |
5. What do central values indicate in data handling? |
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