Ganit Prakash - Class-X - Statistics: Ogive
Let us work out 26.3 Measures of Central Tendency
📘 Exercise 26.3 Solutions (Q1 - Q4)
Chapter 26The following distribution table shows the daily profit (in `) of 100 shops of our village.
| Profit of each shop (in `) | 0-50 | 50-100 | 100-150 | 150-200 | 200-250 | 250-300 |
|---|---|---|---|---|---|---|
| Number of Shops | 10 | 16 | 28 | 22 | 18 | 6 |
Making cumulative frequency table (less than type) of given frequency table, let us draw Ogive on graph paper.
Explanation:
To draw a "less than type" Ogive, we first need to construct the cumulative frequency table. We take the upper class limits and find the cumulative frequency.
| Profit (Upper Class Limit) | Cumulative Frequency (less than type) | Points to Plot $(x, y)$ |
|---|---|---|
| less than 50 | 10 | $(50, 10)$ |
| less than 100 | $10 + 16 = 26$ | $(100, 26)$ |
| less than 150 | $26 + 28 = 54$ | $(150, 54)$ |
| less than 200 | $54 + 22 = 76$ | $(200, 76)$ |
| less than 250 | $76 + 18 = 94$ | $(250, 94)$ |
| less than 300 | $94 + 6 = 100$ | $(300, 100)$ |
Steps to draw the Ogive on Graph Paper:
- Take the upper limits of the class intervals along the X-axis (Profit in `).
- Take the corresponding cumulative frequencies (less than type) along the Y-axis (Number of shops).
- Plot the points $(50, 10)$, $(100, 26)$, $(150, 54)$, $(200, 76)$, $(250, 94)$, and $(300, 100)$.
- Join these points successively by a smooth freehand curve. The curve obtained is the required less than type Ogive.
The following data shows the weight of 35 students of the class of Nivedita.
| Weight (kg.) | less than 38 | less than 40 | less than 42 | less than 44 | less than 46 | less than 48 | less than 50 | less than 52 |
|---|---|---|---|---|---|---|---|---|
| Number of students | 0 | 4 | 6 | 9 | 12 | 28 | 32 | 35 |
Making frequency (less than type) distribution table, let us draw Ogive on graph paper and hence let us find the median from the graph. Let us find the median by using formula and verify it.
Explanation:
Part 1: Drawing the Ogive and finding median from graph
The given table is already a "less than type" cumulative frequency table. The points to be plotted on the graph are:
$(38, 0)$, $(40, 4)$, $(42, 6)$, $(44, 9)$, $(46, 12)$, $(48, 28)$, $(50, 32)$, $(52, 35)$.
Here, $N = 35$, so $\frac{N}{2} = 17.5$. To find the median from the graph, locate $17.5$ on the Y-axis, draw a horizontal line to intersect the Ogive, and from the point of intersection, draw a perpendicular to the X-axis. The X-coordinate is the median (approx. $46.7$ kg).
Part 2: Verifying with Formula
First, we convert the cumulative frequency table into a normal frequency distribution table.
| Weight (kg) | Number of students ($f$) | Cumulative Frequency ($cf$) |
|---|---|---|
| less than 38 | 0 | 0 |
| 38 - 40 | $4 - 0 = 4$ | 4 |
| 40 - 42 | $6 - 4 = 2$ | 6 |
| 42 - 44 | $9 - 6 = 3$ | 9 |
| 44 - 46 | $12 - 9 = 3$ | 12 |
| 46 - 48 | $28 - 12 = 16$ | 28 |
| 48 - 50 | $32 - 28 = 4$ | 32 |
| 50 - 52 | $35 - 32 = 3$ | 35 |
Here, $N = 35 \implies \frac{N}{2} = 17.5$.
The cumulative frequency just greater than $17.5$ is $28$, which corresponds to the median class $46 - 48$.
- Lower limit of median class ($l$) = $46$
- Cumulative frequency of preceding class ($cf$) = $12$
- Frequency of median class ($f$) = $16$
- Class size ($h$) = $2$
The median calculated using the formula verifies the value obtained from the Ogive.
Making cumulative frequency (greater than type) distribution table of given data, let us draw Ogive on graph paper.
| Class | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 |
|---|---|---|---|---|---|---|
| Frequency | 4 | 10 | 15 | 8 | 3 | 5 |
Explanation:
To draw a "greater than type" Ogive, we construct a cumulative frequency table using the lower class limits. The total frequency $N = 4 + 10 + 15 + 8 + 3 + 5 = 45$.
| Class (Lower Limit) | Cumulative Frequency (greater than or equal to type) | Points to Plot $(x, y)$ |
|---|---|---|
| More than or equal to 0 | 45 | $(0, 45)$ |
| More than or equal to 5 | $45 - 4 = 41$ | $(5, 41)$ |
| More than or equal to 10 | $41 - 10 = 31$ | $(10, 31)$ |
| More than or equal to 15 | $31 - 15 = 16$ | $(15, 16)$ |
| More than or equal to 20 | $16 - 8 = 8$ | $(20, 8)$ |
| More than or equal to 25 | $8 - 3 = 5$ | $(25, 5)$ |
Steps to draw the Ogive on Graph Paper:
- Take the lower limits of the class intervals along the X-axis.
- Take the corresponding cumulative frequencies (greater than type) along the Y-axis.
- Plot the points $(0, 45)$, $(5, 41)$, $(10, 31)$, $(15, 16)$, $(20, 8)$, and $(25, 5)$.
- Join these points successively by a smooth freehand curve. The curve obtained is the required greater than type Ogive.
| Class | 100-120 | 120-140 | 140-160 | 160-180 | 180-200 |
|---|---|---|---|---|---|
| Frequency | 12 | 14 | 8 | 6 | 10 |
Drawing less than type Ogive and greater than type Ogive of given data along same axes, on the graph paper, let us find the median from the graph.
Explanation:
Total frequency $N = 12 + 14 + 8 + 6 + 10 = 50$. We need to construct both "less than" and "greater than" cumulative frequency tables.
1. Less than type cumulative frequency table:
| Upper Class Limit | Cumulative Frequency (Less than) | Points to Plot |
|---|---|---|
| Less than 120 | 12 | $(120, 12)$ |
| Less than 140 | $12 + 14 = 26$ | $(140, 26)$ |
| Less than 160 | $26 + 8 = 34$ | $(160, 34)$ |
| Less than 180 | $34 + 6 = 40$ | $(180, 40)$ |
| Less than 200 | $40 + 10 = 50$ | $(200, 50)$ |
2. Greater than type cumulative frequency table:
| Lower Class Limit | Cumulative Frequency (Greater than or equal to) | Points to Plot |
|---|---|---|
| More than or equal to 100 | 50 | $(100, 50)$ |
| More than or equal to 120 | $50 - 12 = 38$ | $(120, 38)$ |
| More than or equal to 140 | $38 - 14 = 24$ | $(140, 24)$ |
| More than or equal to 160 | $24 - 8 = 16$ | $(160, 16)$ |
| More than or equal to 180 | $16 - 6 = 10$ | $(180, 10)$ |
Steps to find Median:
- Plot the "less than" Ogive using the first set of points and join them.
- Plot the "greater than" Ogive using the second set of points on the same axes and join them.
- Mark the point of intersection of the two Ogives.
- Draw a perpendicular from the intersection point to the X-axis. The X-coordinate of this foot of the perpendicular gives the Median.
Mathematical Verification (Optional):
Median class is $120-140$ because $N/2 = 25$ and the CF just greater is $26$.
\[\begin{array}{l} \text{Median} = 120 + \left( \frac{25 - 12}{14} \right) \times 20 \\ = 120 + \left( \frac{13}{14} \right) \times 20 \\ = 120 + \frac{130}{7} \\ \approx 120 + 18.57 = \mathbf{138.57} \end{array}\]The graph's intersection point will align with $\sim 138.6$ on the X-axis.
Hi Please, do not Spam in Comments.