Problem a): Find the mean, median, and mode.

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Here are the solutions to the problems.

Problem a): Find the mean, median, and mode.

First, let's list the data from the table:
Marks Scored ( $x$ ): 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Frequency ( $f$ ): 2, 3, 4, 4, 4, 4, 5, 3, 2, 1

The total number of students is $\sum f = 2+3+4+4+4+4+5+3+2+1 = 32$ .
The sum of ( $fx$ ) is $\sum fx = (1 \times 2) + (2 \times 3) + (3 \times 4) + (4 \times 4) + (5 \times 4) + (6 \times 4) + (7 \times 5) + (8 \times 3) + (9 \times 2) + (10 \times 1)$
$\sum fx = 2 + 6 + 12 + 16 + 20 + 24 + 35 + 24 + 18 + 10 = 167$ .

Mean:

Use the formula for the mean of a frequency distribution.
$\text{Mean} = \frac{\sum fx}{\sum f}$
Substitute the calculated values.
$\text{Mean} = \frac{167}{32}$ $\text{Mean} = 5.21875$
The mean is $\boxed{\textbf{5.21875}}$ .
- Median:
Determine the position of the median.
The total number of data points is $N = \sum f = 32$ . Since $N$ is even, the median is the average of the $\frac{N}{2}$ -th and $(\frac{N}{2}+1)$ -th values.
The positions are $\frac{32}{2} = 16$ th and $\frac{32}{2}+1 = 17$ th.
Find the cumulative frequencies to locate the values.
Marks Scored ( $x$ ) Frequency ( $f$ ) Cumulative Frequency (CF)
1 2 2
2 3 5
3 4 9
4 4 13
5 4 17
6 4 21
7 5 26
8 3 29
9 2 31
10 1 32

The 16th value falls in the '4 marks' category (CF up to 13, next is 17). So the 16th value is 4.
The 17th value falls in the '5 marks' category (CF up to 17). So the 17th value is 5.
Calculate the median.
$\text{Median} = \frac{16\text{th value} + 17\text{th value}}{2}$ $\text{Median} = \frac{4+5}{2}$ $\text{Median} = \frac{9}{2}$ $\text{Median} = 4.5$
The median is $\boxed{\textbf{4.5}}$ .
- Mode:
Identify the mark with the highest frequency.
Looking at the frequency column, the highest frequency is 5, which corresponds to a mark of 7.
The mode is $\boxed{\textbf{7}}$ .

Problem b): Percentage of students who scored at least 8 marks.
Find the number of students who scored at least 8 marks.
"At least 8 marks" means students who scored 8, 9, or 10 marks.
Number of students with 8 marks = 3
Number of students with 9 marks = 2
Number of students with 10 marks = 1
Total students with at least 8 marks = $3 + 2 + 1 = 6$ .
Calculate the percentage.
$\text{Percentage} = \frac{\text{Number of students with at least 8 marks}}{\text{Total number of students}} \times 100\%$ $\text{Percentage} = \frac{6}{32} \times 100\%$ $\text{Percentage} = 0.1875 \times 100\%$ $\text{Percentage} = 18.75\%$
The percentage of students who scored at least 8 marks is $\boxed{\textbf{18.75\%}}$ .

Problem a): Find the mean, median, and mode.

Step-by-step solution

Use the formula for the mean of a frequency distribution.

Substitute the calculated values.

Determine the position of the median.

Find the cumulative frequencies to locate the values.

Calculate the median.

Identify the mark with the highest frequency.

Find the number of students who scored at least 8 marks.

Calculate the percentage.

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