Answer :
Certainly! Let's work through the problem step by step.
### Drawing the 'Less than type' Ogive
1. Gather the Data:
We have two sets of data: the number of wickets and the cumulative number of bowlers.
- Number of wickets: \( 15, 30, 45, 60, 75, 90, 105, 120 \)
- Number of bowlers: \( 2, 5, 9, 17, 39, 54, 70, 80 \)
2. Setting up the Plot:
- Plot the number of wickets on the x-axis.
- Plot the cumulative number of bowlers on the y-axis.
3. Plot the Points:
We'll mark the points corresponding to each pair \((number of wickets, number of bowlers)\) using `(x, y)` form:
- (15, 2)
- (30, 5)
- (45, 9)
- (60, 17)
- (75, 39)
- (90, 54)
- (105, 70)
- (120, 80)
4. Join the Points:
Use a smooth curve to connect these points, which forms our 'Less than type' ogive.
### Finding the Median
To find the median from the ogive:
1. Total Number of Bowlers (\(N\)):
The total number of bowlers is \( 80 \).
2. Median Position:
The median position is at \( \frac{N}{2} = \frac{80}{2} = 40 \).
3. Locate Median Class:
Find the class interval where the cumulative frequency just exceeds or includes 40. Here, the class interval is 60-75 wickets.
4. Details of Median Class:
Let's break down the details needed for interpolation:
- Cumulative frequency just before the median class (CF): 17
- Frequency of the median class (f): \( 39 - 17 = 22 \) (the difference between CF at 75 and CF at 60)
- Lower class boundary of the median class (\(L\)): 60
- Class width (\(h\)): \( 75 - 60 = 15 \)
5. Apply the Median Formula:
The formula for calculating the median in a grouped frequency distribution is:
[tex]\[ \text{Median} = L + \left(\frac{\frac{N}{2} - \text{CF}}{f}\right) \times h \][/tex]
Substitute the values:
[tex]\[ \text{Median} = 60 + \left(\frac{40 - 17}{22}\right) \times 15 \][/tex]
Simplify inside the parenthesis:
[tex]\[ = 60 + \left(\frac{23}{22}\right) \times 15 \][/tex]
Calculate the fraction and the multiplication:
[tex]\[ = 60 + (1.045) \times 15 \][/tex]
[tex]\[ = 60 + 15.675 \][/tex]
[tex]\[ = 75.675 \][/tex]
So, the median number of wickets is approximately 75.68.
### Summary:
- Ogive: A plot connecting (15, 2), (30, 5), (45, 9), (60, 17), (75, 39), (90, 54), (105, 70), and (120, 80) with a smooth curve.
- Median: The median number of wickets taken by the bowlers is approximately 75.68 wickets.
### Drawing the 'Less than type' Ogive
1. Gather the Data:
We have two sets of data: the number of wickets and the cumulative number of bowlers.
- Number of wickets: \( 15, 30, 45, 60, 75, 90, 105, 120 \)
- Number of bowlers: \( 2, 5, 9, 17, 39, 54, 70, 80 \)
2. Setting up the Plot:
- Plot the number of wickets on the x-axis.
- Plot the cumulative number of bowlers on the y-axis.
3. Plot the Points:
We'll mark the points corresponding to each pair \((number of wickets, number of bowlers)\) using `(x, y)` form:
- (15, 2)
- (30, 5)
- (45, 9)
- (60, 17)
- (75, 39)
- (90, 54)
- (105, 70)
- (120, 80)
4. Join the Points:
Use a smooth curve to connect these points, which forms our 'Less than type' ogive.
### Finding the Median
To find the median from the ogive:
1. Total Number of Bowlers (\(N\)):
The total number of bowlers is \( 80 \).
2. Median Position:
The median position is at \( \frac{N}{2} = \frac{80}{2} = 40 \).
3. Locate Median Class:
Find the class interval where the cumulative frequency just exceeds or includes 40. Here, the class interval is 60-75 wickets.
4. Details of Median Class:
Let's break down the details needed for interpolation:
- Cumulative frequency just before the median class (CF): 17
- Frequency of the median class (f): \( 39 - 17 = 22 \) (the difference between CF at 75 and CF at 60)
- Lower class boundary of the median class (\(L\)): 60
- Class width (\(h\)): \( 75 - 60 = 15 \)
5. Apply the Median Formula:
The formula for calculating the median in a grouped frequency distribution is:
[tex]\[ \text{Median} = L + \left(\frac{\frac{N}{2} - \text{CF}}{f}\right) \times h \][/tex]
Substitute the values:
[tex]\[ \text{Median} = 60 + \left(\frac{40 - 17}{22}\right) \times 15 \][/tex]
Simplify inside the parenthesis:
[tex]\[ = 60 + \left(\frac{23}{22}\right) \times 15 \][/tex]
Calculate the fraction and the multiplication:
[tex]\[ = 60 + (1.045) \times 15 \][/tex]
[tex]\[ = 60 + 15.675 \][/tex]
[tex]\[ = 75.675 \][/tex]
So, the median number of wickets is approximately 75.68.
### Summary:
- Ogive: A plot connecting (15, 2), (30, 5), (45, 9), (60, 17), (75, 39), (90, 54), (105, 70), and (120, 80) with a smooth curve.
- Median: The median number of wickets taken by the bowlers is approximately 75.68 wickets.
