Random Variables

Lesson

What happens to the mean and variance of a probability distribution when all outcomes experience a change in scale and/or origin?

Let's first experiment with some data.

Let's take the heights of $10$10 students in a class, recorded in metres.

$1.64$1.64 $1.67$1.67 $1.70$1.70 $1.71$1.71 $1.63$1.63 $1.55$1.55 $1.60$1.60 $1.71$1.71 $1.69$1.69 $1.72$1.72

The population mean and standard deviation for this data set is mean = $1.662$1.662 and standard deviation = $0.053$0.053

What would happen to the mean and standard deviation if we had recorded these results in centimetres instead of metres?

To investigate this we need to convert all the measurements to cm, and to do this we need to multiply each score by $100$100. This is called a change of scale.

If you try this yourself you should calculate that the mean = $166.2$166.2 and the standard deviation = $5.3$5.3

So what's happened to the mean and standard deviation? They've both been multiplied by the same scale factor as the data. That is, they've also been multiplied by $100$100.

What if instead, we decided we needed to subtract $2$2 cm from each height (that is, $0.02$0.02 m) to account for the height of the heel of school shoes?

Once again, you will first need to subtract $0.02$0.02 from each original data point and recalculate the mean and standard deviation. This is called a change of origin.

If you try this you should calculate that mean = $1.64$1.642 and the standard deviation = $0.053$0.053

So what's happened to the mean and standard deviation? Only the mean has decreased by $0.02$0.02, the standard deviation has stayed the same.

This makes complete sense. If you add or subtract an amount from all data points you will not change the spread of the data, just the location of the mean, and so the standard deviation won't change.

The same effects as we saw above occur if we change the scale and/or origin of the outcomes in a probability distribution.

Change of Scale and/or Origin on Expected Value and Variance

Let's say have the probability distribution for $X$`X` whose outcomes undergo the transformations $aX+b$`a``X`+`b`, then:

$E(aX+b)=aE(X)+b$`E`(`a``X`+`b`)=`a``E`(`X`)+`b`

$Var(aX+b)=a^2Var(X)$`V``a``r`(`a``X`+`b`)=`a`2`V``a``r`(`X`)

Note that we multiply the variance by $a^2$`a`2, not $a$`a`, since this is variance and not standard deviation.

Consider the probability distribution for a discrete random variable given in the table below.

$x$x |
$1$1 | $2$2 | $3$3 | $4$4 | $5$5 |
---|---|---|---|---|---|

$P(X=x)$P(X=x) |
$0.2$0.2 | $0.15$0.15 | $0.4$0.4 | $0.1$0.1 | $0.15$0.15 |

Calculate:

(a) $E(2X-4)$`E`(2`X`−4)

(b) $Var(2X-4)$`V``a``r`(2`X`−4)

We saw in the last few chapters that $E(X)=2.85$`E`(`X`)=2.85 and $Var(X)=1.6275$`V``a``r`(`X`)=1.6275

To calculate the new expected value under a change of scale of $2$2 and a change of origin of $-4$−4, we apply both of these to our original expected value, since the mean is affected by both.

So $E\left(2X-4\right)=2\times2.85-4$`E`(2`X`−4)=2×2.85−4

$E\left(2X-4\right)=1.7$`E`(2`X`−4)=1.7

To calculate the new variance, it will only be affected by the change of scale, not the change of origin.

So $Var\left(2X-4\right)=2^2\times1.6275$`V``a``r`(2`X`−4)=22×1.6275

$Var\left(2X-4\right)=6.51$`V``a``r`(2`X`−4)=6.51

The expected value of a discrete random variable is $E\left(X\right)=4.3$`E`(`X`)=4.3

Calculate $2E\left(X\right)$2

`E`(`X`).Calculate $E\left(2X\right)$

`E`(2`X`).Calculate $E\left(X-1\right)$

`E`(`X`−1).Calculate $E\left(4X+1\right)$

`E`(4`X`+1).Solve for the value of $n$

`n`such that $E\left(nX+3\right)=11.6$`E`(`n``X`+3)=11.6.

The table below represents the distribution of a discrete random variable, where $E\left(X\right)$`E`(`X`) is the expected value, $V\left(X\right)$`V`(`X`) is the variance and $S\left(X\right)$`S`(`X`) is the standard deviation.

$x$x |
$1$1 | $2$2 | $3$3 | $4$4 | $5$5 |
---|---|---|---|---|---|

$P$P$($($X=x$X=x$)$) |
$0.5$0.5 | $m$m |
$0.15$0.15 | $n$n |
$0.05$0.05 |

Express $m$

`m`in terms of $n$`n`.Given that $E\left(X^2\right)=5.5$

`E`(`X`2)=5.5, solve for the value of $n$`n`.Hence find $m$

`m`.Calculate $E\left(X\right)$

`E`(`X`).Calculate $V\left(X\right)$

`V`(`X`).Calculate $E\left(5X+7\right)$

`E`(5`X`+7).Calculate $S\left(4-X\right)$

`S`(4−`X`).Give your answer to one decimal place.

Spacemaths is selling an online eBook for $\$20$$20 per student.

The table below shows the probabilities associated with the number of students purchasing the product and the associated costs for the company.

Number of students purchasing | Costs | Probability |
---|---|---|

$50000$50000 | $\$70000$$70000 | $0.2$0.2 |

$100000$100000 | $\$100000$$100000 | $0.1$0.1 |

$200000$200000 | $\$120000$$120000 | $0.4$0.4 |

$500000$500000 | $\$150000$$150000 | $0.3$0.3 |

Calculate the expected number of students purchasing the product.

Hence calculate the expected revenue.

Calculate the expected cost for Spacemaths.

Hence calculate the expected profit for Spacemaths.

If students were offered a $25%$25% discount, calculate the expected profit for Spacemaths. Assume the probabilities are the same.

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

Apply probability distributions in solving problems