---
title: "Normal: 2000 students above 60%, top 100"
subject: "Probability and Statistics"
chapter: 3
key: "semester_5/probability_and_statistics:normal-2000-students-above-60-top"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/normal-2000-students-above-60-top
site: BCT Engineering
---

# Normal: 2000 students above 60%, top 100

Probability and Statistics · Chapter 3: Continuous Probability Distributions

- Asked in: 2073 Chaitra, 5 marks
- Video: [ Normal Distribution - Probability Distribution - Problem](https://www.youtube.com/watch?v=yHqCaoCBnHM) — Btech Maths Hub
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/normal-2000-students-above-60-top
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-3

## Question

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.

## 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%.

## Answer

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).

```text
 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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For X ~ N(50, 5), what is z for 60% marks?
   A. 2.00
   B. 10.00
   C. 0.50
   D. 1.645
   Answer: A. z = (60 − 50)/5 = 2.00. (Key point: Standardising)

2. P(Z > 2.00) is obtained from the table as:
   A. 0.4772
   B. 0.5 + 0.4772
   C. 0.5 − 0.4772 = 0.0228
   D. 1 − 0.0228
   Answer: C. The table gives area from 0 to z; the right tail is 0.5 minus that, so 0.0228, giving about 46 of 2000 students. (Key point: Tail area to count)

3. How is the cut-off for the top 100 of 2000 students set?
   A. P(X > x0) = 0.5
   B. P(X > x0) = 0.05, so table area 0.45 and z = 1.645
   C. P(X > x0) = 0.10, so z = 1.28
   D. P(X > x0) = 0.95, so z = −1.645
   Answer: B. 100/2000 = 0.05 lies to the right of x0, so area from 0 to z is 0.45, z = 1.645. (Key point: Top 100 fraction)

4. Using z = 1.645, the minimum marks of the top 100 are:
   A. 50 − 1.645 × 5
   B. 50 + 1.645 × 5 = 58.2
   C. 50 + 1.645/5
   D. 1.645 × 50
   Answer: B. x0 = μ + zσ = 50 + 8.225 = 58.2 approximately. (Key point: Back-transform)

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