Topic 4.15 · HL only

Linear transformation of random variable(s), unbiased estimate of mean and var: notes and practice questions

Summary

An estimator is a random variable; an estimate is its numerical value from a sample.
An estimator is unbiased (HL) if its expected value equals the population parameter: $E(\text{Estimator}) = \text{Parameter}$ .
The sample mean ( $\bar{x}$ ) is always an unbiased estimate for the population mean ( $\mu$ ).
The standard sample variance ( $s_n^2$ ) is not an unbiased estimate for population variance ( $\sigma^2$ ) (HL); it systematically underestimates.
A linear combination of independent normal variables is also normally distributed (HL).
For a linear transformation of a single random variable (HL):
$E(aX + b) = aE(X) + b$
$\text{Var}(aX + b) = a^2\text{Var}(X)$
Adding a constant ( $b$ ) does not affect variance.
For linear combinations of multiple independent random variables (HL):
$E(a_1X_1 \pm \dots \pm a_nX_n) = a_1E(X_1) \pm \dots \pm a_nE(X_n)$
$\text{Var}(a_1X_1 \pm \dots \pm a_nX_n) = a_1^2\text{Var}(X_1) + \dots + a_n^2\text{Var}(X_n)$
Always add variances, even when subtracting variables.
Unbiased estimate of population mean (HL): $\bar{x} = \frac{\sum x}{n}$ .
Unbiased estimate of population variance (HL): $s_{n-1}^2 = \frac{n}{n-1} s_n^2$ .
Expected value of the sample mean (HL): $E(\bar{X}) = \mu$ .
Variance of the sample mean (HL): $\text{Var}(\bar{X}) = \frac{\sigma^2}{n}$ .
Expected value of the sample mean squared (HL): $E(\bar{X}^2) = \mu^2 + \frac{\sigma^2}{n}$ .
When modeling contextual sums (HL), differentiate scaling a single variable (e.g., $6E$ ) from summing multiple independent variables (e.g., $E_1 + \dots + E_6$ ).
HL syllabus explicitly covers algebraic expected value/variance formulas and unbiased estimates.
GDC notation for unbiased standard deviation is often $s_x$ or $s_{n-1}$ ; square it for unbiased variance.
Never subtract variances; $\text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y)$ .
"Sample variance" refers to $s_n^2$ ; adjust with $\frac{n}{n-1}$ for the unbiased estimate.

How it is examined

The $a^2$ in the variance transformation is the fact students most often get wrong, and it is a one-mark answer. Distinguishing $s_n$ from $s_{n-1}$ on the GDC is the practical skill, since the calculator reports both and only one is the unbiased estimate. Proofs of unbiasedness are not examinable, so a question can only ask for the values.

Given in the booklet

The unbiased estimate of the population variance, $s_{n-1}^2$ .

Key ideas

Linear transformation of a single random variable.
The expected value of linear combinations of $n$ random variables.
The variance of linear combinations of $n$ independent random variables.
$\bar{x}$ as an unbiased estimate of $\mu$ .

Not assessed

$\text{Var}(X)$ is the variance of the random variable $X$ . The variance formula will not be required in examinations.
**Demonstration that $E(\bar{X}) = \mu$ and $E(s_{n-1}^2) = \sigma^2$ will not be examined**, but may help understanding.

Linking questions

TOK: mathematics and the world. In the absence of knowing the value of a parameter, will an unbiased estimator always be better than a biased one?

Practice questions

6 questions · 1 medium · 5 hard

Showing 6 of 6

Question 1

MediumPaper 1 · calculator6 marks

(a)

A materials scientist is developing a new synthetic fiber and claims its average tensile strength is 120 N. To test this claim, she takes a random sample of 25 fiber segments.

Given that the tensile strength of an individual fiber segment is $x$ newtons, the scientist found that, from her 25 samples, $\sum x = 2968$ and $\sum x^2 = 352480$ .

(a) Find an unbiased estimate for the mean tensile strength ( $\mu$ ) of the fiber.

[1]

(b)

(b) Use the formula $s_{n-1}^2 = \frac{\sum x^2 - \frac{(\sum x)^2}{n}}{n-1}$ to determine an unbiased estimate for the variance of the tensile strength of the fiber.

[2]

(c)

(c) Find a 95% confidence interval for $\mu$ . You may assume that all conditions for a confidence interval have been met.

[2]

(d)

(d) Suggest, with justification, a valid conclusion that the materials scientist could make regarding her claim.

[1]

Question 2

HardPaper 2 · calculator19 marks

(a)

A game involves a player attempting to hit a target. The number of successful hits, $X$ , in a round follows a discrete probability distribution given by:

$X$

$0$

$1$

$2$

$3$

$P(X=x)$

$\frac{1}{8}$

$\frac{3}{8}$

$\frac{1}{8}$

(a) Find $P(X \le 1)$ .

[1]

(b)(i)

(b) Giving your answers as fractions, calculate:

(i) $E(X)$

[3]

(c)(i)

(i) Find $E(Y)$ .

[3]

(c)(ii)

(ii) Find $Var(Y)$ .

[2]

(d)(i)

(d) Let the random variable $T = X_1 + X_2 + X_3 + X_4$ , be the total of four independent values of $X$ .

(i) Find $E(T)$ .

[3]

(d)(ii)

(ii) Find $Var(T)$ .

[2]

(e)(i)

(e) Let the random variable $R = \frac{1}{X+1}$ .

(i) Calculate $E(R)$ giving the answer as a fraction.

[3]

(e)(ii)

(ii) Hence, determine whether or not the statement $E\left(\frac{1}{X+1}\right) = \frac{1}{E(X)+1}$ is true. You must justify your answer.

[2]

Question 3

HardPaper 2 · calculator12 marks

(a)

(a) A logistics company handles two types of packages: small (S) and large (L). The weight of each type of package follows a normal distribution with parameters as shown in this table:

Type of Package

Mean weight (kg)

Standard deviation (kg)

S (Small)

$1.5$

$0.2$

L (Large)

$5.0$

$0.5$

One package of each type is selected at random. Find the probability that the large package weighs less than four times the weight of the small package.

[5]

(b)

(b) One large package and three small packages are selected at random. Find the probability that the large package weighs more than the total weight of the three small packages.

[7]

Question 4

HardPaper 2 · calculator16 marks

(a)

A factory produces specialized electronic components. The total "quality score" of a component, $T$ , is a combination of scores from three independent inspection stages:

Stage 1: Automated visual inspection. The score $X$ from this stage has an expectation of $2.5$ and a standard deviation of $0.7$ .

Stage 2: Manual functional test. A batch of $8$ critical functions are tested, and the score $Y$ is the number of functions that pass. Each function has a $0.35$ probability of passing, independently.

Stage 3: Environmental stress test. The score $Z$ from this stage, representing the number of successful stress cycles, follows a Poisson distribution with a mean of $4.2$ .

The overall quality score for a component is given by $T = X + 2Y + Z$ .

Calculate the expected value and variance of the total quality score $T$ .

[6]

(b)

Given that the distribution of $T$ can be approximated by a Normal distribution, find the probability that a randomly selected component has a total quality score between $10$ and $14$ (inclusive of $10$ , exclusive of $14$ ).

[4]

(c)

The factory manager wants to ensure that the mean total quality score of a sample of $n$ components is within $0.8$ units of the true mean, with a probability of at least $0.95$ . Find the minimum sample size $n$ required.

[6]

Question 5

HardPaper 2 · calculator15 marks

(a)

A quality control manager at a manufacturing plant wants to assess the consistency of a new batch of electronic components. He decides to test ten components, ensuring that five are selected from Production Line A and five from Production Line B. The manager instructs the supervisors of each line to provide the required number of components from their current production.

(a) Name the type of sampling that best describes the method used by the quality control manager.

[1]

(b)(i)

The weights, in grams, of the ten components selected for the test are:

$148, 153, 161, 155, 142, 160, 150, 157, 145, 159$ .

(b) For these ten components, find

(i) the mean weight.

[2]

(b)(ii)

(ii) the standard deviation of the weights.

[2]

(c)

The target weight for these components is $155$ g. The manager is concerned that the components might be consistently underweight. Perform an appropriate test at the $10\%$ significance level to see if the mean weight of the components produced is less than the target weight. It can be assumed that the weights come from a normal population.

[5]

(d)

State one reason why the test performed in part (c) might not be valid.

[1]

(e)(i)

Two additional components are tested at a later date. The mean weight for all twelve components is $154.5$ g and the standard deviation is $7.2$ g.

For further analysis, a 'quality score' for the twelve components is obtained by multiplying the weights by $1.5$ and subtracting $50$ .

(e) For the twelve components, find

(i) their mean quality score.

[2]

(e)(ii)

(ii) the standard deviation of their quality score.

[2]

Question 6

HardPaper 2 · calculator18 marks

(a)

The battery life, in hours, of a particular smartphone model, $L$ , can be modelled by a normal distribution with a mean of 24 hours and a standard deviation of 2 hours.

(a) Find the probability that a randomly selected smartphone has a battery life greater than 27 hours.

[2]

(b)(i)

Two smartphones are selected at random and independently of each other.

(b) (i) Find the probability that both smartphones have a battery life greater than 27 hours.

[2]

(b)(ii)

(b) (ii) Find the probability that their total battery life is greater than 52 hours.

[4]

(c)

A software update is released which is claimed to improve battery life. The manufacturer decides to take a random sample of 20 smartphones to test this claim at the 1% significance level, assuming the standard deviation of the battery life has not changed.

[1]

(d)

(d) Find the critical region for this test.

[4]

(e)

Unknown to the manufacturer, the software update has resulted in all smartphones having a 5% longer battery life than the original model.

(e) Find the mean and standard deviation of the battery life for smartphones with the update.

[3]

(f)

(f) Find the probability of a Type II error in the manufacturer’s test.

[2]

Every Linear transformation of random variable(s), unbiased estimate of mean and var question, marked for you

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Where marks are lost

Answering to the wrong accuracy. Two significant figures, or six, where the rule says exactly or three. Common wherever a GDC's full decimal display gets copied straight down.
Rounding an intermediate value and then using it in a later part. Costs a mark every time, and AI's multi-part modelling questions give it more chances to happen than AA's shorter, more self-contained ones.
Writing the answer and nothing else, where the mark scheme has an explicit M1 rather than an implied one. A bare answer cannot score full marks there.

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What does Linear transformation of random variable(s), unbiased estimate of mean and var cover in IB Maths AI?

An estimator is a random variable; an estimate is its numerical value from a sample. An estimator is unbiased (HL) if its expected value equals the population parameter: E(Estimator) = Parameter. The sample mean (barx) is always an unbiased estimate for the population mean (μ).

Is Linear transformation of random variable(s), unbiased estimate of mean and var SL or HL?

Linear transformation of random variable(s), unbiased estimate of mean and var is HL only. SL students are not examined on it.

How do I revise Linear transformation of random variable(s), unbiased estimate of mean and var for IB Maths AI?

Start from the core idea: an estimator is a random variable; an estimate is its numerical value from a sample. In the exam: the a^2 in the variance transformation is the fact students most often get wrong, and it is a one-mark answer. Distinguishing s_n from s_n-1 on the GDC is the practical skill, since the calculator reports both and only one is the unbiased estimate. Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

How does FourtyFive help me practise Linear transformation of random variable(s), unbiased estimate of mean and var?

FourtyFive has 6 Linear transformation of random variable(s), unbiased estimate of mean and var questions. Every answer you write is marked mark by mark, IB-style, and you see where each mark was won or lost. Every part has a hint, the AI tutor helps you through the step you are stuck on, and your Study Profile picks what to practise next.

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Linear transformation of random variable(s), unbiased estimate of mean and var: notes and practice questions

How it is examined

Linking questions

Practice questions

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Where marks are lost

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How it is examined

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Question 6

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Where marks are lost

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