In the previous lesson, Transforming Random Variables I, we considered the following problem, and partially solved it, but did not yet know how to handle the $0.30 per customer materials cost. In this lesson we will learn how to multiply random variables by a constant (like 0.30), so we can finish solving the problem.

Suppose you are running a lemonade stand, and you know that you have a mean of 15 customers per hour, with a standard deviation of 5 customers. If you know that it costs you 30 cents per glass in materials, and that you could hire your little sister to work the stand for you for $4.50 per hour, how could you calculate the mean and standard deviation of the cost per hour of having your sister run the stand?

### Transforming Random Variables (Part 2)

We discussed previously the process of adding or subtracting a constant to a random variable, but often, you may need to multiply the random variable by a constant as well, or instead. For instance, suppose you had a random variable that represented the temperature in your classroom in degrees Celsius, and wanted to convert the values to degrees Fahrenheit. You could either start over with your calculations, having first converted each random variable outcome to the Fahrenheit equivalent, or you could just convert the mean and variance you already calculated to Fahrenheit by multiplying by \begin{align*}\frac{9}{5}\end{align*}

Since we discussed the process of adding a constant to a random variable in the previous lesson, let’s look at the multiplication process.

The graph below represents the probability distribution of random variable \begin{align*}Y\end{align*}

Let’s see what happens if we multiply each outcome by 3, and find \begin{align*}\mu_{3Y}\end{align*}

By multiplying every outcome by three, not only was the mean multiplied by three: \begin{align*}\mu_{3Y} = 3 \times 3=9\end{align*}^{2} or 9.

#### Transforming Variables

1. Given that \begin{align*}\mu_X=3\end{align*}

When a random variable is multiplied by a constant, the mean and standard deviation are multiplied by the same constant, and the variance is multiplied by the square of the constant:

- \begin{align*}\mu_{3X}=\mu_X \times 3= 3 \times 3=9\end{align*}
μ3X=μX×3=3×3=9 - \begin{align*}\sigma_{3X}=\sigma_X \times 3= .5 \times 3 = 1.5\end{align*}
σ3X=σX×3=.5×3=1.5 - \begin{align*}{\sigma^2}_{3X}={\sigma^2}_X \times 3^2=.25 \times 9=2.25\end{align*}
σ23X=σ2X×32=.25×9=2.25

2. Given random variable \begin{align*}Y\end{align*}

As above, when a random variable is multiplied by a constant, the mean and standard deviation are multiplied by the same constant, and the variance is multiplied by the square of the constant:

- \begin{align*}\mu_{-2.2Y} = \mu_Y \times -2.2 = 0.75 \times -2.2 = -1.65\end{align*}
- \begin{align*}\sigma_{-2.2y} = \sigma_Y \times -2.2 = 0.125 \times -2.2 = -0.275\end{align*}
- \begin{align*}{\sigma^2}_{-2.2Y} = {\sigma^2}_Y \times (-2.2)^2 = 0.0156 \times 4.84 = 0.075\end{align*}

#### Finding the Mean and Standard Deviation

Tony the troll charges a fee of 3 gold pieces per person + 2 gold pieces per sword in order to cross his bridge. If random variable *S* represents the number of swords carried by a traveler wanting to cross his bridge, then \begin{align*}\mu_S=2.8\end{align*}, and \begin{align*}\sigma_S=1.095\end{align*} What is the mean and standard deviation of Tony’s income per traveler

Since we know the mean and variance of the traffic across Tony’s bridge, we are really just looking to multiply that data by 2 and then add 3 in order to convert the number of swords to the number of gold pieces and add the per traveler charge.

We can describe the converted values mathematically, using the random variable *S* from the question, as: \begin{align*}\mu_{2S+3}\end{align*}, and \begin{align*}\sigma_{2S+3}\end{align*}.

We know that adding a constant affects the mean, but not the variance or standard deviation, and that multiplying by a constant affects both, so the standard deviation should be multiplied by two, and the mean should be multiplied by 2, and then increased by 3:

- \begin{align*}\mu_{2S+3} = 2({\mu_S})+3 = 2(2.8)+3=5.6+3=8.6\end{align*}
- \begin{align*}\sigma_{2S+3} = 2(\sigma_S)=2(1.095)=2.19\end{align*} (remember that adding a constant to a random variable doesn’t affect the standard deviation.)

#### Earlier Problem Revisited

*Suppose you are running a lemonade stand, and you know that you have a mean of 15 customers per hour, with a standard deviation of 5 customers. If you know that it costs you 30 cents per glass in materials, and that you could hire your little sister to work the stand for you for $4.50 per hour, how could you calculate the mean and standard deviation of the cost per hour of having your sister run the stand?*

Now we should have all of the pieces to solve this problem. If we let random variable \begin{align*}D\end{align*} equal the number of customers per hour, then \begin{align*}\mu_D=15\end{align*} and \begin{align*}\sigma_D=5\end{align*}. We need to multiply the number of customers, \begin{align*}D\end{align*}, by $0.30, and add that to the base cost of $4.50 that we have to pay our sister regardless of the number of customers, to get the hourly cost in dollars. Mathematically, we need to find the mean and standard deviation of \begin{align*}.3D + 4.5\end{align*}. In other words, we are looking for \begin{align*}\mu_{.3D+4.5}\end{align*} and \begin{align*}\sigma_{.3D+4.5}\end{align*}.

- \begin{align*}\mu_{.3D+4.5}=.3(\mu_D)+4.5=.3(15)+4.5=4.5+4.5= \$ 9.00 \ per \ hour\end{align*}
- \begin{align*}\sigma_{.3D+4.5}=.3(\sigma_D)=.3(5)=\$ 1.50 \ per \ hour\end{align*}

So, based on our mean number of customers, and paying our sister $4.50 per hour, we can expect to have a total of $9.00 per hour in materials and labor, with a standard deviation of $1.50.

### Examples

#### Example 1

If \begin{align*}\mu_X=5.5\end{align*}, and \begin{align*}\sigma_X=.8\end{align*}, what is \begin{align*}\mu_{2.5X}\end{align*} and \begin{align*}\sigma_{2.5X}\end{align*}?

Multiplying a random variable by a constant affects both mean and standard deviation:

- \begin{align*}\mu_{2.5X}=(2.5)(\mu_X)=(2.5)(5.5)=13.75\end{align*}
- \begin{align*}\sigma_{2.5X}=(2.5)(\sigma_X)=(2.5)(0.8)=.2\end{align*}

#### Example 2

Given \begin{align*}\mu_Y=3.1\end{align*}, \begin{align*}\sigma_X=.35\end{align*}, and \begin{align*}{\sigma^2}_X=.1225\end{align*}, what would \begin{align*}\mu_{-1.3X}\end{align*}, \begin{align*}\sigma_{-1.3X}\end{align*} and \begin{align*}{\sigma^2}_{-1.3X}\end{align*} be?

As above, in \begin{align*}Q1\end{align*}:

- \begin{align*}\mu_{-1.3X}=(-1.3)(\mu_X)=(-1.3)(3.1)=-4.03\end{align*}
- \begin{align*}\sigma_{-1.3X}=(-1.3)(\sigma_X)=(-1.3)(.35)=-0.455\end{align*}
- \begin{align*}{\sigma^2}_{-1.3X}=(-1.3)^2({\sigma^2}_X)=(1.69)(0.1225)=0.207\end{align*}

#### Example 3

If random variable \begin{align*}Z\end{align*} has \begin{align*}\mu=14.9\end{align*}, and \begin{align*}\sigma^2=16\end{align*}, what is the mean and the standard deviation of \begin{align*}4Z+2\end{align*}?

Multiplying by a constant affects both mean and variance, adding a constant affects only the mean, so here we get:

- \begin{align*}\mu_{4Z+2}=4(\mu_Z)+2=4(14.9)+2=59.6+2=61.6\end{align*}
- \begin{align*}{\sigma^2}_{4Z+2}=4^2({\sigma^2}Z)=(16)(16)=256\end{align*}

### Review

- If \begin{align*}\mu_X=.23\end{align*}, and \begin{align*}\sigma_X=.03\end{align*}, what is \begin{align*}\mu_{3.1X}\end{align*} and \begin{align*}\sigma_{3.1X}\end{align*}?
- If \begin{align*}\mu_C=124\end{align*}, and \begin{align*}\sigma_C=12\end{align*}, what is \begin{align*}\mu_{4C}\end{align*} and \begin{align*}\sigma_{4C}\end{align*}?
- If \begin{align*}\mu_Y=19.2\end{align*}, and \begin{align*}\sigma_Y=2.3\end{align*}, what is \begin{align*}\mu_{-1.9Y}\end{align*} and \begin{align*}\sigma_{-1.9Y}\end{align*}?
- If \begin{align*}\mu_Z=48.38\end{align*}, and \begin{align*}\sigma_Z=2.27\end{align*}, what is the \begin{align*}\mu\end{align*} and \begin{align*}\sigma\end{align*} of \begin{align*}2.3Z\end{align*}?
- If \begin{align*}\mu_X=7.84\end{align*}, and \begin{align*}\sigma_X=.72\end{align*}, what is the \begin{align*}\mu\end{align*} and \begin{align*}\sigma\end{align*} of \begin{align*}-1.98X\end{align*}?
- If \begin{align*}\mu_A=\frac{5}{7}\end{align*}, and \begin{align*}\sigma_A=\frac{1}{7}\end{align*}, what is the \begin{align*}\mu\end{align*} and \begin{align*}\sigma\end{align*} of \begin{align*}\frac{3}{14}A\end{align*}?
- If \begin{align*}\mu_B=14.11\end{align*}, and \begin{align*}\sigma_B=1.15\end{align*}, what is the \begin{align*}\mu\end{align*} and \begin{align*}\sigma\end{align*} of \begin{align*}-0.35B+2\end{align*}?
- If \begin{align*}\mu_A=\frac{3}{7}\end{align*}, and \begin{align*}\sigma_A=\frac{1}{7}\end{align*}, what is the \begin{align*}\mu\end{align*} and \begin{align*}\sigma\end{align*} of \begin{align*}2A -\frac{3}{14}\end{align*}?
- If \begin{align*}\mu_Y=21.3\end{align*}, and \begin{align*}\sigma_Y=2.94\end{align*}, what is the \begin{align*}\mu\end{align*} and \begin{align*}\sigma\end{align*} of \begin{align*}-3.8Y+2.13\end{align*}?
- If \begin{align*}+1SD\end{align*}
*of*\begin{align*}\mu_X=4.75\end{align*} and \begin{align*}-1SD\end{align*}*of*\begin{align*}\mu_X=3.25\end{align*}, what is \begin{align*}\mu_{3X+2.25}\end{align*} and \begin{align*}\sigma_{3X+2.25}\end{align*}? - If \begin{align*}+1SD\end{align*}
*of*\begin{align*}\mu_X=11.25\end{align*} and \begin{align*}-1SD\end{align*}*of*\begin{align*}\mu_X=9.75\end{align*}, what is \begin{align*}\mu_{2X-2.25}\end{align*} and \begin{align*}\sigma_{2X-2.25}\end{align*}? - If \begin{align*}+2SD\end{align*}
*of*\begin{align*}\mu_X=25\end{align*} and \begin{align*}-2SD\end{align*}*of*\begin{align*}\mu_X=5\end{align*}, what is \begin{align*}\mu_{3X-4}\end{align*} and \begin{align*}\sigma_{3X-4}\end{align*}? - If \begin{align*}-2SD\end{align*}
*of*\begin{align*}\mu_Z=9\end{align*} and \begin{align*}+2SD\end{align*}*of*\begin{align*}\mu_Z=15\end{align*}, what is \begin{align*}\mu_{2Z-.75}\end{align*} and \begin{align*}\sigma_{2Z-.75}\end{align*}?

### Review (Answers)

To view the Review answers, open this PDF file and look for section 7.10.