Consider the following data for the next two (02) items that follow : Class 0-30 30-60 60-90 90-120 Frequency 4 5 7 4
If the median (P) and mode (Q) satisfy the relation 7(Q - P) = 9R, then what is the value of R?
6
This problem requires us to calculate the median (P) and the mode (Q) from the given grouped frequency distribution data and then use these values in a given relationship to find the value of R.
First, let's look at the provided data table:
| Class | 0-30 | 30-60 | 60-90 | 90-120 |
| Frequency | 4 | 5 | 7 | 4 |
Calculating the Median (P) for Grouped Data
The median is the middle value of a dataset. For grouped data, we first find the median class and then use a specific formula.
Steps to find the median:
- Calculate the cumulative frequency (CF).
- Find the total frequency, N.
- Determine the median position, which is \( \frac{N}{2} \).
- Identify the median class, which is the class where the cumulative frequency is just greater than or equal to \( \frac{N}{2} \).
- Use the median formula for grouped data: \( P = L + \frac{\frac{N}{2} - CF}{f} \times h \)
Let's add the cumulative frequency to the table concept:
| Class | Frequency (f) | Cumulative Frequency (CF) |
| 0-30 | 4 | 4 |
| 30-60 | 5 | 4 + 5 = 9 |
| 60-90 | 7 | 9 + 7 = 16 |
| 90-120 | 4 | 16 + 4 = 20 |
Total frequency \( N = 20 \).
Median position \( = \frac{N}{2} = \frac{20}{2} = 10 \).
The cumulative frequency just greater than 10 is 16, which corresponds to the class 60-90. So, the median class is 60-90.
Now, let's use the formula with the values for the median class:
- L (lower limit of median class) = 60
- \( \frac{N}{2} \) = 10
- CF (cumulative frequency of the class preceding the median class) = 9
- f (frequency of the median class) = 7
- h (class width) = 30 - 0 = 30 (or 60 - 30, etc.)
Substitute these values into the median formula:
\( P = 60 + \frac{10 - 9}{7} \times 30 \)
\( P = 60 + \frac{1}{7} \times 30 \)
\( P = 60 + \frac{30}{7} \)
\( P = \frac{60 \times 7 + 30}{7} \)
\( P = \frac{420 + 30}{7} \)
\( P = \frac{450}{7} \)
So, the median \( P = \frac{450}{7} \).
Calculating the Mode (Q) for Grouped Data
The mode is the value that appears most frequently in a dataset. For grouped data, we first find the modal class and then use a specific formula.
Steps to find the mode:
- Identify the modal class, which is the class with the highest frequency.
- Use the mode formula for grouped data: \( Q = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h \)
Looking at the frequency column in the original table, the highest frequency is 7, which corresponds to the class 60-90. So, the modal class is 60-90.
Now, let's use the formula with the values for the modal class:
- L (lower limit of modal class) = 60
- \( f_1 \) (frequency of the modal class) = 7
- \( f_0 \) (frequency of the class preceding the modal class) = 5 (frequency of 30-60 class)
- \( f_2 \) (frequency of the class succeeding the modal class) = 4 (frequency of 90-120 class)
- h (class width) = 30
Substitute these values into the mode formula:
\( Q = 60 + \frac{7 - 5}{2(7) - 5 - 4} \times 30 \)
\( Q = 60 + \frac{2}{14 - 9} \times 30 \)
\( Q = 60 + \frac{2}{5} \times 30 \)
\( Q = 60 + 2 \times 6 \)
\( Q = 60 + 12 \)
\( Q = 72 \)
So, the mode \( Q = 72 \).
Finding the Value of R
The problem gives the relation \( 7(Q - P) = 9R \).
We have found \( P = \frac{450}{7} \) and \( Q = 72 \).
Substitute these values into the equation:
\( 7 \left( 72 - \frac{450}{7} \right) = 9R \)
To simplify the expression inside the parenthesis, find a common denominator:
\( 72 - \frac{450}{7} = \frac{72 \times 7}{7} - \frac{450}{7} = \frac{504}{7} - \frac{450}{7} = \frac{504 - 450}{7} = \frac{54}{7} \)
Now substitute this back into the relation:
\( 7 \left( \frac{54}{7} \right) = 9R \)
The 7 in the numerator and the 7 in the denominator cancel out:
\( 54 = 9R \)
To find R, divide both sides by 9:
\( R = \frac{54}{9} \)
\( R = 6 \)
Thus, the value of R is 6.
Revision Table: Formulas for Grouped Data
| Measure | Formula | Notes |
| Median (P) | \( P = L + \frac{\frac{N}{2} - CF}{f} \times h \) | L = lower limit of median class, N = total frequency, CF = cumulative frequency of preceding class, f = frequency of median class, h = class width |
| Mode (Q) | \( Q = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h \) | L = lower limit of modal class, \(f_1\) = freq. of modal class, \(f_0\) = freq. of preceding class, \(f_2\) = freq. of succeeding class, h = class width |
Additional Information: Measures of Central Tendency
Median and mode are important measures of central tendency, which describe the center of a dataset. For grouped data, their calculation involves specific formulas because we don't have the exact values of each observation.
- Median: Represents the middle value. It's less affected by extreme values than the mean. For grouped data, it's estimated based on its position in the cumulative frequency distribution.
- Mode: Represents the most frequent value. For grouped data, it's the class with the highest frequency (modal class), and the exact value is estimated using the formula that considers the frequencies of the adjacent classes.
- Other measures include the Mean, which is the average value. Calculating the mean for grouped data involves using the class marks (midpoints of the class intervals).
Understanding how to calculate these measures from grouped frequency distributions is fundamental in statistics for summarizing and interpreting data.
Similar Questions
The sum of deviations of n numbers from 10 and 20 are a, b respectively. If \(\frac{b}{a}\) = -4, then what is the mean of these n numbers ?
What is the mode of the distribution?
What is the median of the following data?
2, 3, -1, 2, 6, 8, 9
What is the arithmetic mean of the first ten composite numbers?
The ages of 7 family members are 2, 5, 12, 18, 38, 40 and 60 years respectively. After 5 years a new member aged x years is added. If the mean age of the family now goes up by 1.5 years, then what is the value of x?
What is the value of p ?
What is the value of q ?
what is the median of the distribution ?
- The sum of deviations of a set of n values measured from 50 is – 10 and the sum of deviations of the values measured from 46 is 70. What is the mean of the values ?
What is the mean of frequency distribution of Series-I?
Important Questions from Elementary Statistics
In a colony 5 families have 1 child, 7 families have 2 children, 8 families have 3 children and 3 families have 4 children.What is the mode of the number of children.
What will be the difference between mean and median of the given data?
21, 11, 27, 8, 5, 12, 7, 23, 3, 14, 9, 19Find the mode and median of 3, 4, 5, 5, 3, 6, 7, 3, 5, 5, 6.
A. 5 and 5
B. 3 and 5
C. 5 and 4
D. 3 and 4
For which set of numbers do the mean, median and mode all have the same value?
The median of 5, 8, 25, 22, 34, 18 is
