Class 10 Maths 1 Statistics Practice Set 6.2 Solutions | Maharashtra Board
Class 10 Mathematics 1 – Statistics
Practice Set 6.2 – Complete Solutions
Welcome to Rukmini Classes!
In this article, we provide complete and detailed solutions to Class 10 Mathematics 1, Chapter 6: Statistics, Practice Set 6.2, according to the Maharashtra State Board syllabus.
In this practice set, students will learn how to calculate the median of grouped frequency distributions using the appropriate formula. All four questions are explained step by step, including cumulative frequency tables, identification of the median class, substitution of values in the formula, and final answers.
These solutions are presented in a simple and easy-to-understand manner to help students understand the concept of median, improve their mathematical skills, and prepare effectively for the SSC Board Examination.
Read each solution carefully and practise the calculations to gain confidence in solving Statistics problems.
Let us begin with the formula for finding the median.
Class 10 Mathematics 1 – Chapter 6: Statistics
Practice Set 6.2 – Complete Solutions
Welcome to Rukmini Classes! In this article, we provide detailed, step-by-step solutions to Practice Set 6.2 of Class 10 Mathematics 1 (Statistics), according to the Maharashtra State Board syllabus.
Formula for Median
To find the median of a grouped frequency distribution, use the following formula:
Median = L + [(N/2 - F) / f] × h
Where:
- L = Lower limit of the median class
- N = Total frequency
- F = Cumulative frequency before the median class
- f = Frequency of the median class
- h = Class interval width
Question 1
The following table shows the classification of the number of workers and the number of hours they work in a software company. Find the median of the number of hours they work.
| Daily Number of Hours | Number of Workers (f) | Cumulative Frequency (cf) |
|---|---|---|
| 8–10 | 150 | 150 |
| 10–12 | 500 | 650 |
| 12–14 | 300 | 950 |
| 14–16 | 50 | 1000 |
| Total | 1000 |
Step 1: Find the total frequency.
N = 150 + 500 + 300 + 50
N = 1000
N/2 = 1000/2 = 500
Step 2: Identify the median class.
The cumulative frequency just greater than or equal to 500 is 650. Therefore, the median class is 10–12.
Here,
- L = 10
- N = 1000
- F = 150
- f = 500
- h = 12 − 10 = 2
Step 3: Apply the formula.
Median = L + [(N/2 − F) / f] × h
Median = 10 + [(500 − 150) / 500] × 2
Median = 10 + (350/500) × 2
Median = 10 + 0.7 × 2
Median = 10 + 1.4
Median = 11.4 hours
Answer: The median number of working hours is 11.4 hours.
Question 2
The frequency distribution table shows the number of mango trees in a grove and their yield of mangoes. Find the median of the data.
| Number of Mangoes | Number of Trees (f) | Cumulative Frequency (cf) |
|---|---|---|
| 50–100 | 33 | 33 |
| 100–150 | 30 | 63 |
| 150–200 | 90 | 153 |
| 200–250 | 80 | 233 |
| 250–300 | 17 | 250 |
| Total | 250 |
Step 1: Find the total frequency.
N = 33 + 30 + 90 + 80 + 17
N = 250
N/2 = 250/2 = 125
Step 2: Identify the median class.
The cumulative frequency just greater than or equal to 125 is 153. Therefore, the median class is 150–200.
Here,
- L = 150
- N = 250
- F = 63
- f = 90
- h = 200 − 150 = 50
Step 3: Apply the formula.
Median = L + [(N/2 − F) / f] × h
Median = 150 + [(125 − 63) / 90] × 50
Median = 150 + (62/90) × 50
Median = 150 + 34.44
Median ≈ 184.44 mangoes
Answer: The median yield is approximately 184.44 mangoes.
Question 3
The following table shows the classification of the number of vehicles and their speeds on the Mumbai–Pune Expressway. Find the median of the data.
| Average Speed of Vehicles (km/hr) | Number of Vehicles (f) | Cumulative Frequency (cf) |
|---|---|---|
| 60–64 | 10 | 10 |
| 64–69 | 34 | 44 |
| 70–74 | 55 | 99 |
| 75–79 | 85 | 184 |
| 79–84 | 10 | 194 |
| 84–89 | 6 | 200 |
| Total | 200 |
Step 1: Find the total frequency.
N = 10 + 34 + 55 + 85 + 10 + 6
N = 200
N/2 = 200/2 = 100
Step 2: Identify the median class.
The cumulative frequency just greater than or equal to 100 is 184. Therefore, the median class is 75–79.
Assuming continuous class boundaries, the values are:
- L = 74.5
- N = 200
- F = 99
- f = 85
- h = 5
Step 3: Apply the formula.
Median = L + [(N/2 − F) / f] × h
Median = 74.5 + [(100 − 99) / 85] × 5
Median = 74.5 + (1/85) × 5
Median = 74.5 + 0.0588
Median ≈ 74.56 km/hr
Answer: Assuming the median class has a lower boundary of 74.5, the median speed is approximately 74.56 km/hr.
Note: The speed intervals printed in the question are not perfectly continuous. The calculation above uses the assumed continuous boundary for the 75–79 class.
Question 4
The production of electric bulbs in different factories is shown in the following table. Find the median of the productions.
| Number of Bulbs Produced (Thousands) | Number of Factories (f) | Cumulative Frequency (cf) |
|---|---|---|
| 30–40 | 12 | 12 |
| 40–50 | 35 | 47 |
| 50–60 | 20 | 67 |
| 60–70 | 15 | 82 |
| 70–80 | 8 | 90 |
| 80–90 | 7 | 97 |
| 90–100 | 8 | 105 |
| Total | 105 |
Step 1: Find the total frequency.
N = 12 + 35 + 20 + 15 + 8 + 7 + 8
N = 105
N/2 = 105/2 = 52.5
Step 2: Identify the median class.
The cumulative frequency just greater than or equal to 52.5 is 67. Therefore, the median class is 50–60.
Here,
- L = 50
- N = 105
- F = 47
- f = 20
- h = 60 − 50 = 10
Step 3: Apply the formula.
Median = L + [(N/2 − F) / f] × h
Median = 50 + [(52.5 − 47) / 20] × 10
Median = 50 + (5.5/20) × 10
Median = 50 + 2.75
Median = 52.75 thousand bulbs
Answer: The median production is 52.75 thousand bulbs, or 52,750 bulbs.
Final Answers
| Question | Median |
|---|---|
| Q.1 | 11.4 hours |
| Q.2 | Approximately 184.44 mangoes |
| Q.3 | Approximately 74.56 km/hr (under the stated boundary assumption) |
| Q.4 | 52.75 thousand bulbs |
Thank you for visiting Rukmini Classes. Practise these solutions to understand the median formula and prepare for your Maharashtra State Board SSC examination.
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