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Central tendency and dispersion; median, pass marks

Short answer

Median marks = 36.67. Minimum pass marks for 60% passing (40th percentile) = 33.33. Convert ‘more than’ data to frequencies: 5,10,15,10,6,3,1. Median class 30–40, P₄₀ also in 30–40.

Define measures of central tendency and measures of variance. Following data gives the distribution of marks of 50 students in statistics.

Marks more than10203040506070
No. of students504535201041
Compute median marks. Also compute minimum marks obtained by a pass candidate if 60% student pass in the test.

Measures of central tendency

These give one representative value around which the data cluster. Examples: mean, median, mode.

Measures of variance (dispersion)

These show how widely the values are spread about the centre. Examples: range, quartile deviation, mean deviation, variance and standard deviation.

Convert "more than" into a frequency table

Number in a class = (more than lower limit) − (more than next limit). The last class is taken as 70-80.

Marksfcf (less than)
10-2050 − 45 = 55
20-3045 − 35 = 1015
30-4035 − 20 = 1530
40-5020 − 10 = 1040
50-6010 − 4 = 646
60-704 − 1 = 349
70-80150

Median

N/2 = 25, so the median class is 30-40 (cf before = 15, f = 15).

Median = 30 + (25 − 15)/15 × 10 = 30 + 6.67 = 36.67 marks

Pass mark

If 60% pass, the lowest 40% fail. The pass mark is the 40th percentile (P₄₀): position 0.40 × 50 = 20, class 30-40.

P40 = 30 + (20 − 15)/15 × 10 = 30 + 3.33 = 33.33 marks

Answer: median = 36.67 marks; minimum marks of a passing candidate ≈ 33.3 (about 34 in whole marks).

✓ Answer checked by Claude Sonnet 5.5

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