NZ Level 8 (NZC) Level 3 (NCEA) [In development]
topic badge
Linear change of scale and origin
Lesson

Consider a set of measurements $x_1,x_2,x_3,...,x_n$x1,x2,x3,...,xn They could be students' scores achieved in a test or the lengths of pieces of string or many other things. Whatever their physical origin, we can think of them as values of a random variable $X$X.

We calculate the mean of $X$X by summing the measurements and dividing by the number of them. In symbols, we write

$\mu_X=\frac{1}{n}\sum_{i=1}^nx_i$μX=1nni=1xi.

We calculate the variance of $X$X by summing the squared distances of the measurements from the mean and dividing by the number of them. In symbols, we write

$\text{Var}X=\frac{1}{n}\sum_{i=1}^n(x_i-\mu_X)^2$VarX=1nni=1(xiμX)2

 

Change of origin

Suppose we added $20$20 points to every student's test score or decreased the length of every piece of string by $10$10cm or generally, added a fixed amount a to every $x_i$xi to obtain a new random variable $Y$Y. We could write $Y=X+a$Y=X+a.

mean

Now, the mean of $Y$Y is $\mu_Y=\frac{1}{n}\sum_{i=1}^n(x_i+a)$μY=1nni=1(xi+a) since $a$a has been added to every measurement. Thus, 

$\mu_Y$μY $=$= $\frac{1}{n}\left(\sum_{i=1}^nx_i+na\right)$1n(ni=1xi+na)
  $=$= $\mu_X+a$μX+a

This is not unexpected. If the value of every observation is shifted by a fixed amount, then the mean should also move by that amount. 

variance

The variance of $Y$Y must be 

$\text{Var}Y$VarY $=$= $\frac{1}{n}\sum_{i=1}^n\left(y_i-\mu_Y\right)^2$1nni=1(yiμY)2
  $=$= $\frac{1}{n}\sum_{i=1}^n\left(x_i+a-(\mu_X+a)\right)^2$1nni=1(xi+a(μX+a))2
  $=$= $\text{Var}X$VarX

Thus, adding a constant to every measurement does not affect the variance.

 

Scaling

Suppose we scaled the scores of the students by a factor of $1.3$1.3 or we decided to express the lengths of the pieces of string in metres rather than centimetres or generally, we multiplied every $x_i$xi by a number $b$b to obtain a new random variable $Z=bX$Z=bX.

mean

The mean of $Z$Z must be

$\mu_Z$μZ $=$= $\frac{1}{n}\sum_{i=1}^nbx_i$1nni=1bxi
  $=$= $\frac{b}{n}\sum_{i=1}^nx_i$bnni=1xi
  $=$= $b\mu_X$bμX

Again, this is not unexpected. If the value of every observation is multiplied by a fixed amount, then the mean should also be multiplied by that amount. 

variance

The variance of Z must be

$\text{Var}Z$VarZ $=$= $\frac{1}{n}\sum_{i=1}^n(z_i-\mu_Z)^2$1nni=1(ziμZ)2
  $=$= $\frac{1}{n}\sum_{i=1}^n(bx_i-b\mu_X)^2$1nni=1(bxibμX)2
  $=$= $\frac{b^2}{n}\sum_{i=1}^n(x_i-\mu_X)^2$b2nni=1(xiμX)2
  $=$= $b^2\text{Var}X$b2VarX

 

 

Putting all of this together, we see that if $X$X and $W$W are random variables with $W$W a linear transformation of $X$X, we have

$W=aX+b$W=aX+b

and

$\mu_W=a\mu_X+b$μW=aμX+b

$\text{Var}W=a^2\text{Var}X$VarW=a2VarX

 

Example

A school mathematics test was given marks out of $50$50. It was found that some of the questions on the test were very easy and every student was able to answer them correctly. The easy questions were worth $14$14 marks and the teacher decided to remove those marks from all the scores so that the test was now effectively marked out of $36$36. The school required that the test results be recorded as a mark out of $100$100 so that every score now had to be scaled up by a factor of $\frac{100}{36}$10036.

If the original mean was $33$33 and the original variance was $9$9, what were the mean and standard deviation reported in the school's records?

The original scores can be represented as a random variable $X$X and the final transformed scores by $Y$Y. The transformation is given by $Y=\frac{25}{9}(X-14)$Y=259(X14). That is, $Y=\frac{25X}{9}-\frac{350}{9}$Y=25X93509.

We must have, for the mean, $\mu_Y=\frac{25}{9}\mu_X-\frac{350}{9}=\frac{25}{9}\times33-\frac{350}{9}=52.7$μY=259μX3509=259×333509=52.7.

For the variance, $\text{Var}Y=\left(\frac{25}{9}\right)^2\text{Var}X=\left(\frac{25}{9}\right)^2\times9=\frac{625}{9}$VarY=(259)2VarX=(259)2×9=6259.

The standard deviation is the square root of the variance. So, the standard deviation of $Y$Y is $\frac{25}{3}\approx8.3$2538.3.

 

 

You will often find another notation for the variance of a random variable. Instead of $\text{Var}X$VarX, we write $\sigma_X^2$σ2X. Then, the standard deviation is $\sigma_X$σX.

You will also find the mean of a random variable referred to as its expected value and so, $\mu_X$μX and $E(X)$E(X) are alternatives.

 

Worked Examples

question 1

The heights of a certain species of fully grown plants are thought to be normally distributed with a mean of $55$55 cm and a standard deviation of $4$4 cm.

If the heights were recorded in mm instead of cm:

  1. State the new mean.

  2. State the new variance.

Question 2

In a given population, a certain variable $X$X is considered to be normally distributed with a mean of $80$80 and a standard deviation of $4$4.

If the data for $Y$Y is transformed according to the rule $-8-4X$84X:

  1. Calculate the new mean.

  2. Calculate the new standard deviation.

Question 3

The marks in the Chemistry ATAR exam were normally distributed. Let $X$X$%$% be the random variable representing the distribution of these marks.

Dylan scored a raw mark of $57%$57% and after average marks scaling scored $63.28%$63.28%.

Danielle scored a raw mark of $86%$86% and after average marks scaling scored $93.44%$93.44%.

  1. If these marks were scaled according to the rule $aX+b$aX+b, determine the values of $a$a and $b$b.

  2. If the raw mean mark was $62%$62%, determine the scaled mean mark for Chemistry.

 

Outcomes

S8-4

Investigate situations that involve elements of chance: A calculating probabilities of independent, combined, and conditional events B calculating and interpreting expected values and standard deviations of discrete random variables C applying distributions such as the Poisson, binomial, and normal

91586

Apply probability distributions in solving problems

What is Mathspace

About Mathspace