Marks 0 1 2 3 4 5 Frequency 7 4 18 12 8 11 The table gives the distribution of marks obtained by a number of…

Analyze the question: We need the median mark from a frequency table.

Concept: The median is the middle value when all the marks are arranged in order. For a frequency table, we use cumulative frequency to find which mark sits at the middle position.

• If total frequency N is even, the median is the average of the (N/2)th and (N/2 + 1)th values.
• If N is odd, the median is the ((N+1)/2)th value.

Solution:

Total frequency: 7 + 4 + 18 + 12 + 8 + 11 = 60

Cumulative frequencies:

| Mark | Frequency | Cumulative Frequency | |------|-----------|----------------------| | 0 | 7 | 7 | | 1 | 4 | 11 | | 2 | 18 | 29 | | 3 | 12 | 41 | | 4 | 8 | 49 | | 5 | 11 | 60 |

N = 60 is even, so the median is the average of the 30th and 31st values.

• Positions 12 to 29 are mark 2.
• Positions 30 to 41 are mark 3.

So the 30th value is 3 and the 31st value is 3.

Median = (3 + 3) / 2 = 3

Answer: The median is 3, which is option B.

Tip: Always build the cumulative frequency column first. Then find the position N/2 (and N/2 + 1 if N is even) and see which mark's cumulative range contains it. Here, the 29th value is the last 2, so the 30th and 31st values both fall in the mark 3 group. Notice how close the 30th position is to the boundary, so be careful with your addition!