---
title: "Normal: IOE marks of 5000 students"
subject: "Probability and Statistics"
chapter: 3
key: "semester_5/probability_and_statistics:normal-5000-students-marks"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/normal-5000-students-marks
site: BCT Engineering
---

# Normal: IOE marks of 5000 students

Probability and Statistics · Chapter 3: Continuous Probability Distributions

- Asked in: 2070 Asar, 4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/normal-5000-students-marks
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-3

## Question

The marks obtained by IOE students in statistics are 50 on average with variance 16. If 5000 students have given the exam, find the following:
a) The number of students securing marks less than 40?
b) The number of students securing marks between 35 to 60?

## Short answer

Standardize: z=(x−50)/4. For marks<40, z=−2.5, probability 0.0062, so about 31 students. For marks between 35 and 60, z=−3.75 and 2.5, probability 0.9937, so about 4969 students.

## Answer

Marks X ~ N(μ = 50, σ² = 16), so σ = 4. Number of students N = 5000.

#### a) Marks less than 40
z = (40 − 50)/4 = −2.50

P(X < 40) = P(Z < −2.50) = 0.5 − 0.4938 = 0.0062

Number = 5000 × 0.0062 = 31 students.

#### b) Marks between 35 and 60
- z₁ = (35 − 50)/4 = −3.75
- z₂ = (60 − 50)/4 = 2.50

P(35 < X < 60) = P(−3.75 < Z < 2.50) = P(0 < Z < 3.75) + P(0 < Z < 2.50) = 0.4999 + 0.4938 = 0.9937

Number = 5000 × 0.9937 = 4969 students.

**Answer: a) about 31 students  b) about 4969 students.**

## Check yourself

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

1. The variance is 16, so the z-score for marks 40 (mean 50) is:
   A. −10/16 = −0.625
   B. −10/4 = −2.5
   C. −10/2 = −5
   D. −2.5 with σ = 16
   Answer: B. σ = sqrt(16) = 4, so z = (40 − 50)/4 = −2.5. (Key point: Standard deviation from variance)

2. P(Z < −2.5) is:
   A. 0.4938
   B. 0.5 − 0.4938 = 0.0062
   C. 0.5 + 0.4938
   D. 1 − 0.0062
   Answer: B. The tail is 0.5 minus the table area, giving 0.0062, so about 31 students of 5000. (Key point: Left tail)

3. For 35 < X < 60, z runs from −3.75 to 2.5. The probability is:
   A. 0.4999 − 0.4938
   B. 0.4999 + 0.4938 = 0.9937
   C. 0.4938 − 0.4999
   D. 0.5 + 0.4938
   Answer: B. When the interval straddles the mean, the two table areas are added. (Key point: Interval spanning the mean)

4. How is the expected number of students obtained?
   A. Probability + 5000
   B. Probability × 5000
   C. Probability / 5000
   D. z × 5000
   Answer: B. Expected count = N × probability, e.g. 5000 × 0.9937 ≈ 4969. (Key point: Converting to counts)

## Related in Probability and Statistics

- [Normal: 2000 students above 60%, top 100](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Anormal-2000-students-above-60-top) — Probability and Statistics, chapter 3
- [Central tendency and dispersion; median, pass marks](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Acentral-tendency-median-pass-marks) — Probability and Statistics, chapter 1
- [CLT: average life of 16 bulbs](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Aclt-bulbs-16-850-750-900) — Probability and Statistics, chapter 4, asked 2×
- [Breakdown voltage: normal, variance 2.25](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Abv-39-42-lt44-gt43) — Probability and Statistics, chapter 3, asked 3×
- [CI: weekly study hours of students](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Astudy-hours-mean-ci-99-95) — Probability and Statistics, chapter 6
- [IOE entrance scores: estimator and consistency](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Aioe-entrance-sixty-estimator-consistency) — Probability and Statistics, chapter 6

## Related in other subjects

- [96 students record; pass marks](https://www.bctengineer.com/question.md?key=semester_1%2Fcomputer_programming%3Afile-96-students-ioe-txt) — Computer Programming, chapter 10
- [3-phase IM principle; speed less than sync](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Athree-phase-im-principle-slip) — Electrical Machine, chapter 5
- [Detection gain of DSB-AM is less than 1](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adetection-gain-dsbam-less-than-1) — Communication Systems, chapter 9
- [When terminal voltage is less or more than emf](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Aterminal-voltage-less-or-more) — Electric Machines II, chapter 1