We appreciate your visit to Question 2 The marks obtained in the previous class test out of 20 were as follows tex begin tabular l l l l l l. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
Sure! Let's solve the problem step by step. We need to organize the data into class intervals and calculate various frequencies, ultimately finding the relative frequency denoted by J in the table.
Step 1: Organize the data into intervals
We're using a class width of 3 and starting the first class at 2. Given the maximum number in our data is 19, our intervals will be:
1. [2, 5)
2. [5, 8)
3. [8, 11)
4. [11, 14)
5. [14, 17)
6. [17, 20)
Step 2: Count the frequencies for each interval
- [2, 5): 2 marks (4 and 5)
- [5, 8): 2 marks (7 and 8)
- [8, 11): 5 marks (9, 9, 10, 10, and 10)
- [11, 14): 7 marks (11, 11, 11, 11, 12, 12, and 12)
- [14, 17): 4 marks (14, 14, 15, and 16)
- [17, 20): 10 marks (17, 17, 18, 18, 19, 19, 19, 19, and 16)
Step 3: Calculate the total number of marks
The total number of marks is the sum of all frequencies: 2 + 2 + 5 + 7 + 4 + 10 = 30.
Step 4: Calculate the relative frequency for each class interval
Relative frequency is the frequency of the interval divided by the total number of marks.
- [2, 5): 2/30 = 0.067
- [5, 8): 2/30 = 0.067
- [8, 11): 5/30 = 0.167
- [11, 14): 7/30 ≈ 0.233
- [14, 17): 4/30 = 0.133
- [17, 20): 10/30 = 0.333
Step 5: Identify the value of J
In the table, J is specified to be the relative frequency value 0.167. This matches the relative frequency calculated for the interval [8, 11).
Therefore, the value of J in the table is indeed 0.167.
Step 1: Organize the data into intervals
We're using a class width of 3 and starting the first class at 2. Given the maximum number in our data is 19, our intervals will be:
1. [2, 5)
2. [5, 8)
3. [8, 11)
4. [11, 14)
5. [14, 17)
6. [17, 20)
Step 2: Count the frequencies for each interval
- [2, 5): 2 marks (4 and 5)
- [5, 8): 2 marks (7 and 8)
- [8, 11): 5 marks (9, 9, 10, 10, and 10)
- [11, 14): 7 marks (11, 11, 11, 11, 12, 12, and 12)
- [14, 17): 4 marks (14, 14, 15, and 16)
- [17, 20): 10 marks (17, 17, 18, 18, 19, 19, 19, 19, and 16)
Step 3: Calculate the total number of marks
The total number of marks is the sum of all frequencies: 2 + 2 + 5 + 7 + 4 + 10 = 30.
Step 4: Calculate the relative frequency for each class interval
Relative frequency is the frequency of the interval divided by the total number of marks.
- [2, 5): 2/30 = 0.067
- [5, 8): 2/30 = 0.067
- [8, 11): 5/30 = 0.167
- [11, 14): 7/30 ≈ 0.233
- [14, 17): 4/30 = 0.133
- [17, 20): 10/30 = 0.333
Step 5: Identify the value of J
In the table, J is specified to be the relative frequency value 0.167. This matches the relative frequency calculated for the interval [8, 11).
Therefore, the value of J in the table is indeed 0.167.
Thanks for taking the time to read Question 2 The marks obtained in the previous class test out of 20 were as follows tex begin tabular l l l l l l. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
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