Measures of Independent Random Variables

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AP Statistics › Measures of Independent Random Variables

Questions 1 - 10

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

$\mu = 20, \sigma = \sqrt{20}$

$\mu = 20, \sigma = 10$

$\mu = 20, \sigma = 40$

$\mu = 10, \sigma = 20$

$\mu = 10, \sigma = 10$

Explanation

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

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

$\mu = 2\cdot 10 = 20$

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

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

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

$\mu = 20, \sigma = \sqrt{20}$

$\mu = 20, \sigma = 10$

$\mu = 20, \sigma = 40$

$\mu = 10, \sigma = 20$

$\mu = 10, \sigma = 10$

Explanation

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

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

$\mu = 2\cdot 10 = 20$

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

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

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:

Explanation

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. $60.5\div10=60.5lbs$

Explanation

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. $60.5\div10=60.5lbs$

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 .

$\mu = 10 + 9 = 19$

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

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 .

$\mu = 10 + 9 = 19$

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

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, $\small \small \frac{X_1+X_2+...+X_{10}}{10}?$

Normal distribution with with mean $\small 0$ and variance $\small \frac{1}{10}$ .

Normal distribution with mean $\small 0$ and variance $\small 1$ .

Normal distribution with mean $\small 0$ and variance $\small 10$ .

Chi-square distribution with $\small 10$ degrees of freedom.

Explanation

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 $\small \frac{1}{10}$ .

Normal distribution with with mean $\small 0$ and variance $\small \frac{1}{10}$ .

Normal distribution with mean $\small 0$ and variance $\small 1$ .

Normal distribution with mean $\small 0$ and variance $\small 10$ .

Chi-square distribution with $\small 10$ degrees of freedom.

Explanation

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 $\small \frac{1}{10}$ .

108.2

110.1

107.2

105.3