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Normal: 2000 students above 60%, top 100

Short answer

Marks are N(50,5). For >60%, z=2 gives P(Z>2)=0.0228, so about 46 students out of 2000. Top 100 correspond to right-tail area 0.05; z=1.645, minimum marks = 50+1.645×5 ≈ 58.23%.

In a certain examination test 2000 students appeared in Statistics. The average marks obtained were 50% and the standard deviation was 5%. How many students do you expect to obtain more than 60% marks? What are the minimum marks of the top 100 students? Assume that the marks are normally distributed.

Let X = marks (%), X ~ N(μ = 50, σ = 5), N = 2000 students.

Students getting more than 60%

z = (x − μ)/σ = (60 − 50)/5 = 2.00

P(X > 60) = P(Z > 2.00) = 0.5 − 0.4772 = 0.0228

Expected number = 2000 × 0.0228 = 45.6, that is about 46 students.

Minimum marks of the top 100 students

Top 100 out of 2000 is a fraction 100/2000 = 0.05 of the students. We need the mark x₀ with P(X > x₀) = 0.05, so the area from 0 to z is 0.5 − 0.05 = 0.45.

From the table, z = 1.645 (area 0.45).

 area 0.05 on the right
            ___
        _.-'   '-._ 
   __.-'     |  '-.__
  ------------+--+---
   50(z=0)  x0 (z=1.645)

x₀ = μ + zσ = 50 + 1.645 × 5 = 58.22

Answer: about 46 students get more than 60%; the top 100 students need at least about 58.2% marks.

✓ Answer checked by Claude Sonnet 5.5

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