How to find the standard deviation of the sum of independent random variables

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AP Statistics › How to find the standard deviation of the sum of independent random variables

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A high school calculus exam is administered to a group of students. Upon grading the exam, it was found that the mean score was 95 with a standard deviation of 12. If one student's z score is 1.10, what is the score that she received on her test?

108.2

110.1

107.2

105.3

109.2

Explanation

The z-score equation is given as: z = (X - μ) / σ, where X is the value of the element, μ is the mean of the population, and σ is the standard deviation. To solve for the student's test score (X):

X = ( z * σ) + 95 = ( 1.10 * 12) + 95 = 108.2.

and are independent random variables. If has a mean of and standard deviation of while variable has a mean of and a standard deviation of , what are the mean and standard deviation of ?

$\mu = 19, \sigma = 10$

$\mu = 19, \sigma = 20$

$\mu = 29, \sigma = 20$

$\mu = 9, \sigma = 10$

$\mu = 15, \sigma = 15$

Explanation

First, find that has $\mu=2\cdot 5 = 10$ and standard deviation $\sigma = 2 \cdot 4 = 8$ .

Then find the mean and standard deviation of .

$\mu = 10 + 9 = 19$

$\sigma = \sqrt{8^2 + 6^2} = \sqrt{100} = 10$

Consider the discrete random variable $\small X$ that takes the following values with the corresponding probabilities:

with $\small \small P(X=1)=\frac{1}{4}$
$\small X=2$ with $\small \small P(X=2)=\frac{1}{4}$
$\small X=3$ with $\small \small P(X=3)=\frac{1}{4}$
$\small X=4$ with $\small P(X=4)=\frac{1}{4}$

Compute the variance of the distribution.

$\small 1.25$

$\small 0.3125$

$\small 1.75$

$\small .625$

Explanation

The variance of a discrete random variable is computed as

$\small Var(X)=\sum_x (x-\mathbb{E}[x])^2P(X=x)$

for all the values of $\small x$ that the random variable $\small X$ can take.

First, we compute $\small \mathbb{E}[X]$ , which is the expected value. In this case, it is $\small 2.5$ .

So we have

$\small \small Var(X)=\frac{1}{4}(1-2.5)^2+\frac{1}{4}(2-2.5)^2+\frac{1}{4}(3-2.5)^2+\frac{1}{4}(4-2.5)^2$

$\small =\frac{1}{4}(2(1.5^2)+2(0.5^2))=\frac{1}{4}(2(2.25)+2(0.25))=\frac{1}{4}(4.5+0.5)$

$\small \small =\frac{5}{4}=1.25$

Clothes 4 Kids uses standard boxes to ship their clothing orders and the mean weight of the clothing packed in the boxes is pounds. The standard deviation is pounds. The mean weight of the boxes is pound with a standard deviation of pounds. The mean weight of the plastic packaging is pounds per box, with a pound standard deviation. What is the standard deviation of the weights of the packed boxes?

$2.021:\mbox{pounds}$

$21.01:\mbox{pounds}$

$4.085:\mbox{pounds}$

$24.00:\mbox{pounds}$

$20.01:\mbox{pounds}$

Explanation

Note that the weight of a packed box = weight of books + weight of box + weight of packing material used.

It is given that $\sigma_{books} = 2: \mbox{pounds}, \sigma_{box} = 0.15: \mbox{pounds}, \mbox{and}: \sigma_{plastic packaging} = 0.25: \mbox{pounds}$ .

The calculation of the standard deviation of the weights of the packed boxes is

$\ \sigma_{packed boxes} = \sqrt{\sigma_{clothes} ^{2} + \sigma_{clothes} ^{2} + \sigma _{plastic packaging}^{2}} \ \ = \sqrt{2^{2}+ 0.15^{2} + 0.25^{2}} \ = 2.021: \mbox{pounds}$

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