Measures of Independent Random Variables - AP Statistics

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If is a random variable with a mean of and standard deviation of , what is the mean and standard deviation of ?

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Remember how the mean and standard deviation of a random variable are affected when it is multiplied by a constant.

$E(cX)=c\cdot E(X)$

$\mu = 2\cdot 10 = 20$

$\sigma = $$\sqrt{2^2$ \cdot 5}$ = $\sqrt{20}$$

If you have ten independent random variables $\small X_1,X_$2,...,X_{10}$$ , normally distributed with mean $\small 0$ and variance $\small 1$ , what is the distribution of the average of the random variables, $\small \small $\frac{X_1+X_$2+...+X_{10}$$}{10}?$

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Any linear combination of independent random variables is also normally distributed with the mean and variance depending on the weights on the random variables. The mean is additive in the sense that

$\small mean\left($\frac{X_$1+...+X_{10}$$}{10}\right)=\frac{1}{10}$\left(mean(X_$1)+...+mean(X_{10}$)\right)$

Each $\small mean(X_i)$ is $\small 0$ , so the sum is equal to zero.

This means the sum of the average

$\small \small \small $\frac{X_1+X_$2+...+X_{10}$$}{10}$ is $\small 0$ .

The variance satisfies

$Var\left($\frac{X_1+X_2+...+X_${10}$}{10}\right)=\frac{1}{10^2$}$Var(X_$1+...+X_{10}$)$

$\small =\frac{1}{100}$(Var(X_$1)+...Var(X_{10}$))=\frac{10}{100}$=\frac{1}{10}$$ because of independence.

This means that the average is normally distributed with mean $\small 0$ and variance $\small $\frac{1}{10}$$ .

Suppose you have three independent normally distributed random variables, $\small X_1,X_2,X_3$ , such that

$\small X_1$ has mean $\small $\frac{1}{4}$$ and variance $\small $\frac{1}{2}$$ ,

$\small X_2$ has mean $\small -1$ and variance $\small $\frac{3}{8}$$ ,

$\small X_3$ has mean $\small $\frac{3}{4}$$ and variance $\small $\frac{1}{8}$$ .

What is the probability that the sum, $\small X_1+X_2+X_3$ , is less than $\small -1$ ?

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There is a relatively simple way of doing this problem. The sum of any set of independent normal random variables is also distributed normally. So $\small X_1+X_2+X_3$ has a normal distribution. Now we can compute the mean and variance. The mean is additive:

$\small mean(X_1+X_2+X_3)=mean(X_1)+mean(X_2)+mean(X_3)$

$\small \small =\frac{1}{4}$-1+\frac{3}{4}$=0$

Variance is also additive in some sense, when the random variables are independent:

$\small \small Var(X_1+X_2+X_3)=Var(X_1)+Var(X_2)+Var(X_3)$

$\small =\frac{1}{2}$+\frac{3}{8}$+\frac{1}{8}$=1$

Thus, $\small X_1+X_2+X_3$ is normally distributed with mean $\small 0$ and variance $\small 1$ .

This sum is a standard normal distribution.

The chance that $\small X_1+X_2+X_3<-1$ is thus $\small 0.1587$ , if we use a normal table.

An experiment is conducted on the watermelons that were grown on a small farm. They want to compare the average weight of the melons grown this year to the average weight of last year's melons. Find the mean of this year's watermelons using the following weights:

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To find the mean you sum up all of your values then divide by the total amount of values. The total sum of the weights is and there are 10 melons. $60.5\div10=60.5lbs$

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?

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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 ?

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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.

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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?

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

It is given that .

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}$

If is a random variable with a mean of and standard deviation of , what is the mean and standard deviation of ?

Tap to see back →

Remember how the mean and standard deviation of a random variable are affected when it is multiplied by a constant.

$E(cX)=c\cdot E(X)$

$\mu = 2\cdot 10 = 20$

$\sigma = $$\sqrt{2^2$ \cdot 5}$ = $\sqrt{20}$$

If you have ten independent random variables $\small X_1,X_$2,...,X_{10}$$ , normally distributed with mean $\small 0$ and variance $\small 1$ , what is the distribution of the average of the random variables, $\small \small $\frac{X_1+X_$2+...+X_{10}$$}{10}?$

Tap to see back →

Any linear combination of independent random variables is also normally distributed with the mean and variance depending on the weights on the random variables. The mean is additive in the sense that

$\small mean\left($\frac{X_$1+...+X_{10}$$}{10}\right)=\frac{1}{10}$\left(mean(X_$1)+...+mean(X_{10}$)\right)$

Each $\small mean(X_i)$ is $\small 0$ , so the sum is equal to zero.

This means the sum of the average

$\small \small \small $\frac{X_1+X_$2+...+X_{10}$$}{10}$ is $\small 0$ .

The variance satisfies

$Var\left($\frac{X_1+X_2+...+X_${10}$}{10}\right)=\frac{1}{10^2$}$Var(X_$1+...+X_{10}$)$

$\small =\frac{1}{100}$(Var(X_$1)+...Var(X_{10}$))=\frac{10}{100}$=\frac{1}{10}$$ because of independence.

This means that the average is normally distributed with mean $\small 0$ and variance $\small $\frac{1}{10}$$ .

Suppose you have three independent normally distributed random variables, $\small X_1,X_2,X_3$ , such that

$\small X_1$ has mean $\small $\frac{1}{4}$$ and variance $\small $\frac{1}{2}$$ ,

$\small X_2$ has mean $\small -1$ and variance $\small $\frac{3}{8}$$ ,

$\small X_3$ has mean $\small $\frac{3}{4}$$ and variance $\small $\frac{1}{8}$$ .

What is the probability that the sum, $\small X_1+X_2+X_3$ , is less than $\small -1$ ?

Tap to see back →

There is a relatively simple way of doing this problem. The sum of any set of independent normal random variables is also distributed normally. So $\small X_1+X_2+X_3$ has a normal distribution. Now we can compute the mean and variance. The mean is additive:

$\small mean(X_1+X_2+X_3)=mean(X_1)+mean(X_2)+mean(X_3)$

$\small \small =\frac{1}{4}$-1+\frac{3}{4}$=0$

Variance is also additive in some sense, when the random variables are independent:

$\small \small Var(X_1+X_2+X_3)=Var(X_1)+Var(X_2)+Var(X_3)$

$\small =\frac{1}{2}$+\frac{3}{8}$+\frac{1}{8}$=1$

Thus, $\small X_1+X_2+X_3$ is normally distributed with mean $\small 0$ and variance $\small 1$ .

This sum is a standard normal distribution.

The chance that $\small X_1+X_2+X_3<-1$ is thus $\small 0.1587$ , if we use a normal table.

An experiment is conducted on the watermelons that were grown on a small farm. They want to compare the average weight of the melons grown this year to the average weight of last year's melons. Find the mean of this year's watermelons using the following weights:

Tap to see back →

To find the mean you sum up all of your values then divide by the total amount of values. The total sum of the weights is and there are 10 melons. $60.5\div10=60.5lbs$

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?

Tap to see back →

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 ?

Tap to see back →

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.

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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?

Tap to see back →

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

It is given that .

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}$

If is a random variable with a mean of and standard deviation of , what is the mean and standard deviation of ?

Tap to see back →

Remember how the mean and standard deviation of a random variable are affected when it is multiplied by a constant.

$E(cX)=c\cdot E(X)$

$\mu = 2\cdot 10 = 20$

$\sigma = $$\sqrt{2^2$ \cdot 5}$ = $\sqrt{20}$$

If you have ten independent random variables $\small X_1,X_$2,...,X_{10}$$ , normally distributed with mean $\small 0$ and variance $\small 1$ , what is the distribution of the average of the random variables, $\small \small $\frac{X_1+X_$2+...+X_{10}$$}{10}?$

Tap to see back →

Any linear combination of independent random variables is also normally distributed with the mean and variance depending on the weights on the random variables. The mean is additive in the sense that

$\small mean\left($\frac{X_$1+...+X_{10}$$}{10}\right)=\frac{1}{10}$\left(mean(X_$1)+...+mean(X_{10}$)\right)$

Each $\small mean(X_i)$ is $\small 0$ , so the sum is equal to zero.

This means the sum of the average

$\small \small \small $\frac{X_1+X_$2+...+X_{10}$$}{10}$ is $\small 0$ .

The variance satisfies

$Var\left($\frac{X_1+X_2+...+X_${10}$}{10}\right)=\frac{1}{10^2$}$Var(X_$1+...+X_{10}$)$

$\small =\frac{1}{100}$(Var(X_$1)+...Var(X_{10}$))=\frac{10}{100}$=\frac{1}{10}$$ because of independence.

This means that the average is normally distributed with mean $\small 0$ and variance $\small $\frac{1}{10}$$ .

Suppose you have three independent normally distributed random variables, $\small X_1,X_2,X_3$ , such that

$\small X_1$ has mean $\small $\frac{1}{4}$$ and variance $\small $\frac{1}{2}$$ ,

$\small X_2$ has mean $\small -1$ and variance $\small $\frac{3}{8}$$ ,

$\small X_3$ has mean $\small $\frac{3}{4}$$ and variance $\small $\frac{1}{8}$$ .

What is the probability that the sum, $\small X_1+X_2+X_3$ , is less than $\small -1$ ?

Tap to see back →

There is a relatively simple way of doing this problem. The sum of any set of independent normal random variables is also distributed normally. So $\small X_1+X_2+X_3$ has a normal distribution. Now we can compute the mean and variance. The mean is additive:

$\small mean(X_1+X_2+X_3)=mean(X_1)+mean(X_2)+mean(X_3)$

$\small \small =\frac{1}{4}$-1+\frac{3}{4}$=0$

Variance is also additive in some sense, when the random variables are independent:

$\small \small Var(X_1+X_2+X_3)=Var(X_1)+Var(X_2)+Var(X_3)$

$\small =\frac{1}{2}$+\frac{3}{8}$+\frac{1}{8}$=1$

Thus, $\small X_1+X_2+X_3$ is normally distributed with mean $\small 0$ and variance $\small 1$ .

This sum is a standard normal distribution.

The chance that $\small X_1+X_2+X_3<-1$ is thus $\small 0.1587$ , if we use a normal table.

An experiment is conducted on the watermelons that were grown on a small farm. They want to compare the average weight of the melons grown this year to the average weight of last year's melons. Find the mean of this year's watermelons using the following weights:

Tap to see back →

To find the mean you sum up all of your values then divide by the total amount of values. The total sum of the weights is and there are 10 melons. $60.5\div10=60.5lbs$