Data on the heights (in cm) of 30 students is given below. Construct a frequency distribution table and find the class interval in which the median height lies. Height class intervals (cm): 145-150, 151-155, 156-160, 161-165 Number of students: 4, 7, 11, 8

hard 4 marks Class 7 · Mathematics · Data Handling

Answer

156-160

The correct answer is 156-160 — Locate the class where cumulative frequency exceeds 15.5: it is $156 - 160$ (since $22 > 15.5$ ).

Step-by-step solution

List the class intervals and corresponding frequencies: $145 - 150 : 4,; 151 - 155 : 7,; 156 - 160 : 11,; 161 - 165 : 8$ (This is the frequency distribution).
Compute cumulative frequencies: $4,; 4 + 7 = 11,; 11 + 11 = 22,; 22 + 8 = 30$ (Cumulative helps locate the median class).
Find the median position: $\frac{N + 1}{2} = \frac{30 + 1}{2} = 15.5$ (For $N = 30$ , the median lies between the 15th and 16th data points).
Locate the class where cumulative frequency exceeds 15.5: it is $156 - 160$ (since $22 > 15.5$ ).

Common mistake: Using the wrong median position formula or not computing cumulative frequencies

Concept tested: Median class interval from frequency distribution