Maharashtra Board Class 10 Math Part – 1 Solution Chapter 6 Practice Set 6.3 – Statistics
Balbharati Maharashtra Board Class 10 Math Part – 1 Solution Chapter 6: Statistics. Marathi or English Medium Students of Class 10 get here Statistics full Exercise Solution.
Std | Maharashtra Class 10 |
Subject | Math Part 1 Solution |
Chapter | Statistics |
Practice Set | 6.3 |
Practice Set 6.3
(1) The following table shows the information regarding the milk collected from farmers on a milk collection centre and the content of fat in the milk, measured by a lactometer. Find the mode of fat content.
Content of fat (%) | 2 – 3 | 3 – 4 | 4 – 5 | 5 – 6 | 6 – 7 |
Milk collected (Litre) | 30 | 70 | 80 | 60 | 20 |
Solution:
From the given table,
The maximum no. of milk collected is from the 4 – 5
Class hence it is the model class.
∴ f1 = 80, f0 = 70, f2 = 60
L = 4, h = 5 – 4 = 1
∴ Mode = L + [f1 – f0/2f1 – f0 – f2] × h
= 4 + [80 – 70/2×80 – 70 – 60] × 1
= 4 + 10/160 – 130
= 4 + 10/30
= 4 + 0 = 33
= 4.33
∴ The mode of content of text in milk is 4.33 litre
(2) Electricity used by some families is shown in the following table. Find the mode for use of electricity.
Use of electricity (Unit) | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 | 100 – 120 |
No. of families | 13 | 50 | 70 | 100 | 80 | 17 |
Solution:
From the given table,
The maximum use of electricity is from 60 – 80 Class hence it is modal class.
∴ f1 = 100, f0 = 70, f2 = 80
L = 60, h = 80 – 60 = 20
∴ Mode = L + [f1 – f0/2f1 – f0 – f2] × h
= 68 + [100 – 70/2×100 – 70 – 80] × 20
= 60 + 30/50 × 20
= 72
∴ The mode of use of electricity by families is 72 units.
(3) Grouped frequency distribution of supply of milk to hotels and the number of hotels is given in the following table. Find the mode of the supply of milk.
Milk (Litre) | 1 – 3 | 3 – 5 | 5 – 7 | 7 – 9 | 9 – 11 | 11 – 13 |
No. of hotels | 7 | 5 | 15 | 20 | 35 | 18 |
Solution:
From the given table,
The maximum milk supplied to hotel is from 9 – 11 Class hence it is our modal class.
∴ f1 = 35, f0 = 20, f2 = 18
L = 9, h = 11 – 9 = 2
∴ Mode = L + [f1 – f0/2f1 – f0 – f2] × h
= 9 + [95 – 20/2×35-20-18] × 2
= 9 + 15/32 × 2
= 9 + 0.9375
= 9.9375
∴ The mode of milk supplied in hotel is 9.9375
(4) The following frequency distribution table gives the ages of 200 patients treated in a hospital in a week. Find the mode of ages of the patients.
Age (Years) | Less than 5 | 5 – 9 | 10 – 14 | 15 – 19 | 20 – 24 | 25 – 29 |
No. of patients | 38 | 32 | 50 | 36 | 24 | 20 |
Solution:
The given table does not have continuous classes hence we have to represent it as continuous classes.
∴ Differences between the upper and lower limits of two consecutive classes respectively is 10 – 9 = 1
∴ To represent it is continuous classes we have to substract and add lower and upper limit of each class respectively with 1/2 = 0.5
Classes | Continuous classes | Frequency |
75 | 0 – 4.5 | 38 |
5 – 9 | 4.5 – 9.5 | 32 -> f0 |
10 – 14 | 9.5 – 14.5 | 50 -> f1 |
15 – 19 | 14.5 – 19.5 | 36 -> f2 |
20 – 24 | 19.5 – 24.5 | 24 |
25 – 29 | 24.5 – 29.5 | 20 |
∴ Mode = L + [f1 – f0/2f1 – f0 – f2] × h
Where, L = 9.5, h = 14.5 – 9.5 = 5
f1 = 50, f0 = 32, f2 = 36
∴ Mode = 9.5 + [50-32/2×50-32-36] × 5
= 9.5 + 8/324 × 5
= 9.5 + 1.25
= 10.75
∴ Mode of ages of patients
= 10.75 years
Here is your solution of Maharashtra Board Class 10 Math Part – 1 Chapter 6 Statistics Practice Set 6.3
Dear Student, I appreciate your efforts and hard work that you all had put in. Thank you for being concerned with us and I wish you for your continued success.