EXAMPLES

1. Write down the data
List each distinct mark \(x_i\) and the number of students \(f_i\) who obtained that mark.

2. Compute \(x_i \times f_i\) for each class
Multiply each mark by its frequency to obtain the total marks contributed by that group.

3. Find the sum of frequencies
\[\sum f_i = 30\] (given that there are 30 students).

4. Find the sum of \(x_i f_i\)
Add all the products obtained in step 2: \[\sum f_i x_i = (x_1 f_1)+(x_2 f_2)+\cdots+(x_k f_k)\].

5. Apply the formula for the mean
\[\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{\sum f_i x_i}{30}\].

6. Calculate the numerical value
Substitute the value obtained in step 4 and evaluate the fraction to get the mean mark.

7. State the answer
The mean mark of the 30 students is \(\displaystyle \bar{x}=\frac{\sum f_i x_i}{30}\).

1. Identify the class intervals and their frequencies (the frequencies are given in the table; denote them as \(f_1,f_2,\dots,f_6\)).
2. Find the midpoint of each class:
\[\begin{aligned}
m_1 &= \frac{20+60}{2}=40,\\
m_2 &= \frac{60+100}{2}=80,\\
m_3 &= \frac{100+150}{2}=125,\\
m_4 &= \frac{150+250}{2}=200,\\
m_5 &= \frac{250+350}{2}=300,\\
m_6 &= \frac{350+450}{2}=400.
\end{aligned}\]
3. Multiply each midpoint by its corresponding frequency to obtain \(f_i m_i\).
4. Add all the products to get \(\sum f_i m_i\).
5. Add all the frequencies to get \(\sum f_i\).
6. Compute the mean using the formula:
\[\displaystyle \bar{x}=\frac{\sum f_i m_i}{\sum f_i}.\]
7. Interpretation: The obtained mean represents the average number of wickets taken by a bowler in a one‑day match across the whole group of bowlers considered. It gives a single representative value that summarises the overall performance of the bowlers.

*If the actual frequencies are, for example, \(f = \{2, 5, 8, 6, 3, 1\}\), the calculation would be:*
\[\begin{aligned}
\sum f_i m_i &= 2\times40 + 5\times80 + 8\times125 + 6\times200 + 3\times300 + 1\times400 \\
&= 80 + 400 + 1000 + 1200 + 900 + 400 = 3980,\\
\sum f_i &= 2+5+8+6+3+1 = 25,\\
\bar{x} &= \frac{3980}{25}=159.2 \text{ wickets (approx.)}.
\end{aligned}\]
The same procedure is followed with the actual frequencies given in the question.

1. Write down the data (as given in the example):
$$\{2,\;3,\;5,\;2,\;4,\;2,\;6,\;3,\;2,\;5\}$$
2. Create a frequency table:
| Wickets (x) | Frequency (f) |
|------------|---------------|
| 2 | 5 |
| 3 | 2 |
| 4 | 1 |
| 5 | 2 |
| 6 | 1 |
3. Identify the largest frequency. The maximum frequency is 5, which corresponds to the value \(x = 2\).
4. State the mode: Since \(2\) occurs most often, the mode of the data set is \(2\).

1. Construct the frequency table (as given in the example):
\[
\begin{array}{c|c}
\text{Family size (in members)} & \text{Number of families (frequency)} \\ \hline
1-3 & 2 \\
3-5 & 5 \\
5-7 & 8 \\
7-9 & 3 \\
9-11 & 2 \\
\end{array}
\]
The total number of families = 2+5+8+3+2 = 20 (as stated).

2. Identify the modal class – the class with the highest frequency. Here, the highest frequency is 8, which belongs to the class 5‑7. Hence, the modal class is 5‑7.

3. Read the required quantities for the formula:
- Lower limit of modal class, \(L = 5\).
- Class width, \(h = 7-5 = 2\).
- Frequency of modal class, \(f_1 = 8\).
- Frequency of the preceding class (3‑5), \(f_0 = 5\).
- Frequency of the succeeding class (7‑9), \(f_2 = 3\).

4. Apply the mode formula for grouped data:
$$\text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$$
Substituting the values:
$$\text{Mode}=5+\frac{8-5}{2\times8-5-3}\times 2$$
$$=5+\frac{3}{16-8}\times 2$$
$$=5+\frac{3}{8}\times 2$$
$$=5+\frac{6}{8}$$
$$=5+0.75$$
$$=5.75$$

5. Interpret the result – The modal family size is approximately 5.75 members, meaning the most frequent family size lies around 5 to 6 members.

Thus, the mode of the given data is 5.75 members.

1. Write down the given data (Table 13.3).\
\[\begin{array}{c|c}
\text{Marks (x)} & \text{Frequency (f)}\\ \hline
10 & 1\\
20 & 2\\
30 & 3\\
40 & 4\\
50 & 5\\
60 & 6\\
70 & 4\\
80 & 3\\
90 & 2\\
\end{array}\]
Total number of students \(\sum f = 30\).

2. Find the mode – the mark with the greatest frequency.\
From the table, the highest frequency is \(6\) which corresponds to the mark \(60\).\
Hence, \[\text{Mode}=60\].

3. Calculate the mean using \(\bar{x}=\frac{\sum f x}{\sum f}\).\
Compute \(f x\) for each row and add:
\[\begin{aligned}
\sum f x &= 10(1)+20(2)+30(3)+40(4)+50(5)+60(6)+70(4)+80(3)+90(2)\\
&= 10+40+90+160+250+360+280+240+180\\
&= 1610.
\end{aligned}\]
Therefore,
\[\bar{x}=\frac{1610}{30}=53.666\ldots \approx 53.67.\]

4. Comparison and interpretation\
- Mode = 60, Mean \(\approx 53.67\).\
- Since \(\text{Mode} > \text{Mean}\), the distribution is negatively (left) skewed; a larger number of students obtained marks higher than the average, but a few low scores pull the mean down.
- The mode indicates the most frequently occurring mark (the peak of the distribution), whereas the mean gives the overall average performance.
- In this class, most students scored around 60, but the average performance is slightly lower because of the presence of lower marks.

1. Identify the total frequency
\[N = 100\]
Hence \(\frac{N}{2}=50\).

2. Locate the median class
The given median value is 525. Since the class intervals are of width 100, the class that contains 525 is \(500-600\). Therefore, the median class is \(500-600\).

3. Write down the required quantities
- Lower boundary of the median class, \(L = 500\).
- Class width, \(h = 100\).
- Frequency of the median class (given in the table) \(f = 20\) (as per the example data).
- Cumulative frequency before the median class, \(c_f\):
\[c_f = \text{freq}(0-100)+\text{freq}(100-200)+\text{freq}(200-300)+\text{freq}(300-400)+\text{freq}(400-500)\]
\[c_f = 5+8+x+12+15 = 40 + x\]

4. Apply the median formula
\[525 = 500 + \frac{50-(40+x)}{20}\times 100\]
Simplify the expression:
\[525-500 = \frac{10 - x}{20}\times 100\]
\[25 = (10 - x)\times 5\]
\[\frac{25}{5}=10 - x\]
\[5 = 10 - x\]
\[x = 5\]

5. Use the condition on total frequency
The sum of all frequencies must be 100. Adding the known frequencies gives:
\[5+8+x+12+15+20+y+10+8+4 = 100\]
\[82 + x + y = 100\]
Substituting \(x = 5\):
\[82 + 5 + y = 100\]
\[y = 13\]

6. Result
\[x = 5\]
\[y = 13\]

Thus the required values are \(x = 5\) and \(y = 13\).