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# Sums and Differences of Independent Random Variables

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Transforming Random Variables II

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 95\begin{align*}\frac{9}{5}\end{align*} and adding 32.

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 Y\begin{align*}Y\end{align*}, with μY=3\begin{align*}\mu_Y=3\end{align*} and σY=1\begin{align*}\sigma_Y=1\end{align*}:

Let’s see what happens if we multiply each outcome by 3, and find μ3Y\begin{align*}\mu_{3Y}\end{align*} and σ3Y\begin{align*}\sigma_{3Y}\end{align*}:

By multiplying every outcome by three, not only was the mean multiplied by three: μ3Y=3×3=9\begin{align*}\mu_{3Y} = 3 \times 3=9\end{align*}, but the standard deviation was also increased by a multiple of three: σ3Y=1×3=3\begin{align*}\sigma_{3Y}=1 \times 3=3\end{align*}. The variance, since it is the square of the standard deviation, would be multiplied by 32 or 9.

#### Transforming Variables

1. Given that μX=3\begin{align*}\mu_X=3\end{align*}, σX=.5\begin{align*}\sigma_X=.5\end{align*}, and σ2X=.25\begin{align*}{\sigma^2}_X= .25\end{align*}, what is μ3X\begin{align*}\mu_{3X}\end{align*}σ3X\begin{align*}\sigma_{3X}\end{align*} and σ23X\begin{align*}{\sigma^2}_{3X}\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:

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

2. Given random variable Y\begin{align*}Y\end{align*} with μY=.75\begin{align*}\mu_Y=.75\end{align*}σY=0.125\begin{align*}\sigma_Y=0.125\end{align*}, and σ2Y=0.0156\begin{align*}{\sigma^2}_Y=0.0156\end{align*}, what is μ2.2Y\begin{align*}\mu_{-2.2Y}\end{align*}σ2.2Y\begin{align*}\sigma_{-2.2Y}\end{align*} and \begin{align*}{\sigma^2}_{-2.2Y}\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*}, by0.30, 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 sister4.50 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

1. 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*}?
2. 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*}?
3. 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*}?
4. 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*}?
5. 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*}?
6. 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*}?
7. 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*}?
8. 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*}?
9. 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*}?
10. 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*}?
11. 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*}?
12. 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*}?
13. 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*}?

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### Vocabulary Language: English

expected value

The expected value is the return or cost you can expect on average, given many trials.

mean

The mean, often called the average, of a numerical set of data is simply the sum of the data values divided by the number of values.

mean of a linear transformation

the mean of a linear transformation is given by the following: uc+dx = c+dux

random variables

Random variables are quantities that take on different values depending on chance, or probability.

standard deviation of a linear transformation

The standard deviation of a linear transformation has the form: $\sigma_{c+dX} = |d| \sigma_X$

Transformation

The transformation of a random variable describes the effect of performing operations (+, -, *, /) on the random variable.

variance

A measure of the spread of the data set equal to the mean of the squared variations of each data value from the mean of the data set.