The DAV Maths Class 7 Solutions and DAV Class 7 Maths Chapter 12 Worksheet 3 Solutions of Data Handling offer comprehensive answers to textbook questions.
DAV Class 7 Maths Ch 12 WS 3 Solutions
Question 1.
Find the mode of the following observations:
(z) 25, 14, 28, 17, 18, 14, 25, 14, 17, 14
(ii) 0, 6, 1, 5, 6, 2, 0, 4, 6, 3, 4, 6, 3, 1
Answer:
(i) The given observations are
25, 14, 28, 17, 18, 14, 25, 14, 17, 14
Here the observation 14 repeats 4 times
∴ Modal observation = 14
(ii) The given observations are
0, 6,1, 5, 6, 2, 0, 4, 6, 3, 4, 6, 3,1
Here the observation 6 comes 4 times
∴ Modal observation is 6.
Question 2.
The following table shows the weight of 15 students. Find the mode.
| Weight (in kg) | Number of students |
| 46 | 3 |
| 51 | 1 |
| 53 | 5 |
| 56 | 2 |
| 60 | 4 |
Answer:
The frequency of 53 kg is more i.e. 5
Mode is 53 kg.
Question 3.
The following table shows sale of shirts having different sizes from a certain shop in a month. Find the mode.
Answer:
Number of shirts of size 39 is 31, which is maximum frequency.
Hence the Mode is 39.
Question 4.
In January 2015, the number of children in 10 families of a locality are:
4, 3, 4, 0, 2, 2, 5, 2, 1, 3 Find the mean, median and mode.
At the end of the year, two families having children 0 and 1 vacated the house. As a result, two more families having children 2 and 5 got the vacant accommodation. Find the new mean, median and mode.
Answer:
The given observations are 4, 3, 4, 0, 2, 2, 5, 2, 1, 3
Mean = \(\frac{\text { Sum of the observations }}{\text { Number of observations }}\)
= \(\frac{4+3+4+0+2+2+5+2+1+3}{10}=\frac{26}{10}\)
= 2.6
Hence mean = 2.6
Arranging the observations in the increasing order,
0, 1, 2, 2, 2, 3, 3, 4, 4, 5
Number of observations = 10 (even)
Median = \(\frac{\frac{10^{\text {th }}}{2} \text { observation }+\left(\frac{10}{2}+1\right)^{\text {th }} \text { observation }}{2}\)
= \(\frac{5^{\text {th }} \text { observation }+6^{\text {th }} \text { observation }}{2}=\frac{2+3}{2}\)
= 2.5
Hence Median = 2.5
Mode: In the given observations 0,1, 2, 2, 2, 3, 3, 4, 4, 5
2 occurs 3 times the most. Therefore mode = 2
Now in the end of the year, the two observations 0 and 1 have been replaced by 2 and 5 respectively.
∴ New observations are 2, 5, 2, 2, 2, 3, 3,4, 4, 5
New Mean = \(\frac{2+5+2+2+2+3+3+4+4+5}{10}=\frac{32}{10}\) = 3.2
New Median: Arranging the new observations in increasing order, 2, 2, 2, 2, 3, 3, 4, 4, 5, 5
Number of observation is 10(even)
New Median = \(\frac{\frac{10^{\text {th }}}{2} \text { observation }+\left(\frac{10}{2}+1\right)^{\text {th }} \text { observation }}{2}\)
= \(\frac{5^{\text {th }} \text { observation }+6^{\text {th }} \text { observation }}{2}=\frac{3+3}{2}\) = 3
New Mode : Given observations are 2, 2, 2, 2, 3, 3, 4, 4, 5, 5
Here 2 repeats 4 times, the most
∴ Mode = 2.