---
title: "IOE entrance scores: estimator and consistency"
subject: "Probability and Statistics"
chapter: 6
key: "semester_5/probability_and_statistics:ioe-entrance-sixty-estimator-consistency"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/ioe-entrance-sixty-estimator-consistency
site: BCT Engineering
---

# IOE entrance scores: estimator and consistency

Probability and Statistics · Chapter 6: Inference concerning mean

- Asked in: 2074 Chaitra, 8 marks
- Video: [Test for Consistency | Probability and Statistics | Prashant YT | 4th Sem | BE Civil | Engineering |](https://www.youtube.com/watch?v=VHKAvFlUdhU) — Prashant YT
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/ioe-entrance-sixty-estimator-consistency
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6

## Question

Following data reveals the scores of sixty candidates of IOE entrance examination
51.43 40 78.57 46.43 51.43 50.71
42.14 50.71 42.86 55 71.43 64.29
52.86 42.14 57.14 45.71 43.57 40
44.29 55.71 40 48.57 48.57 49.29
51.43 47.14 54.29 45 53.57 50
49.29 60 48.57 50.71 50 49.29
47.14 53.57 58.57 43.57 47.14 53.57
47.86 47.14 40 43.57 52.86 47.86
49.29 49.29 42.86 47.14 48.57 50
47.14 50.71 52.86 47.86 47.14 70
a) Estimate average score of candidates.
b) Find unbiased estimator of true standard deviation and standard error of average score.
c) Also test for consistency of score.

## Short answer

Average score: 50.13. Unbiased standard deviation: 7.33, standard error: 0.946. The sample mean is consistent because it is unbiased and its variance σ²/n → 0 as n grows.

## Answer

n = 60 scores. Sum of scores Σx = 3007.85, Σx² = 153956.05.

#### a) Estimate of the average score

The sample mean is the point estimator of the population mean μ:

x̄ = Σx/n = 3007.85/60 = **50.131**

#### b) Unbiased estimator of σ and standard error of the mean

The sample variance with divisor (n − 1) is an unbiased estimator of σ²:

s² = [Σx² − n·x̄²] / (n − 1) = [153956.05 − 60×2513.100]/59 = 53.729

s = √53.729 = **7.330** (estimate of σ; E(s²) = σ²)

Standard error of the mean:
SE(x̄) = s/√n = 7.330/√60 = 7.330/7.746 = **0.946**

#### c) Consistency

An estimator T is **consistent** for θ if T → θ in probability as n → ∞; sufficient conditions are E(T) → θ and Var(T) → 0.

For the sample mean:
- E(x̄) = μ (unbiased)
- Var(x̄) = σ²/n → 0 as n → ∞

So x̄ is a consistent estimator of μ. Using s = 7.330 as the estimate of σ, the standard error falls as the sample grows:

| n | 15 | 30 | 60 | 240 |
|---|---|---|---|---|
| SE = s/√n | 1.893 | 1.338 | 0.946 | 0.473 |

The spread of x̄ shrinks toward zero, so the average score of the sample becomes a more and more accurate estimate of the true mean; likewise s² is a consistent estimator of σ².

**Answer: x̄ = 50.13, s = 7.33, SE = 0.946; x̄ is unbiased and consistent.**

## Check yourself

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

1. What is the point estimate of the average score and its value?
   A. Sample mean = 3007.85/59 = 50.98
   B. Median, 48.57
   C. Sample mean = 3007.85/60 = 50.13
   D. Mode, 47.14
   Answer: C. The sample mean is the point estimator of mu: 3007.85/60 = 50.131. (Key point: Point estimator of mean)

2. Why is the divisor (n - 1) used in s^2?
   A. It makes s^2 an unbiased estimator of sigma^2
   B. It makes the SD smaller than the range
   C. It makes s^2 consistent only
   D. It gives the standard error directly
   Answer: A. With divisor n-1, E(s^2) = sigma^2, so s^2 is unbiased. (Key point: Unbiased variance)

3. What is the standard error of the mean here?
   A. 7.330/60 = 0.122
   B. 7.330
   C. 7.330/sqrt(60) = 0.946
   D. 7.330 x sqrt(60) = 56.8
   Answer: C. SE = s/sqrt(n) = 7.330/7.746 = 0.946. (Key point: Standard error)

4. Why is the sample mean a consistent estimator of mu?
   A. Because it equals mu in every sample
   B. Because its standard error stays at 0.946
   C. Because its variance increases with n
   D. Because it is unbiased and Var = sigma^2/n tends to 0 as n grows
   Answer: D. E(xbar) = mu and Var(xbar) = sigma^2/n -> 0, so xbar converges to mu in probability. (Key point: Consistency)

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