How to find standard deviation of a random variable - AP Statistics

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We have two independent, normally distributed random variables $\small X_1$ and $\small X_2$ such that $\small X_1$ has mean $\small 0$ and variance $\small 2$ and $\small X_2$ has mean $\small 0$ and variance $\small 1$ . What is the probability distribution of the difference of the random variables, $\small \small X_1-X_2$ ?

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The mean for any set of random variables is additive in the sense that

$\small Mean(X_1+X_2)=Mean(X_1)+Mean(X_2)$

The difference is also additive, so we have

$\small Mean(X_1-X_2)=Mean(X_1)-Mean(X_2)$

This means the mean of $\small X_1-X_2$ is $\small 0$ .

The variance is additive when the random variables are independent, which they are in this case. But it's additive in the sense that for any real numbers $\small a,b$ (even when negative), we have

$\small Var(aX_1+bX_$2)=a^2Var$(X_$1)+b^2Var$(X_2)$ .

So for this difference, we have

$\small \small \small Var(X_1-X_2)=Var(X_1+(-1)X_$2)=1^2Var$(X_$1)+(-1)^2Var$(X_2)$

$\small =Var(X_1)+Var(X_2)=2+1=3$ .

So the mean and variance are $\small 0$ and $\small 3$ , respectively. In addition to that, $\small X_1-X_2$ is normally distributed because the sum or difference of any set of independent normal random variables is also normally distributed.

Robert's work schedule for next week will be released today. Robert will work either 45, 40, 25, or 12 hours. The probabilities for each possibility are listed below:

45 hours: 0.3

40 hours: 0.2

25 hours: 0.4

12 hours: 0.1

What is the standard deviation of the possible outcomes?

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There are four steps to finding the standard deviation of random variables. First, calculate the mean of the random variables. Second, for each value in the group (45, 40, 25, and 12), subtract the mean from each and multiply the result by the probability of that outcome occurring. Third, add the four results together. Fourth, find the square root of the result.

$\sqrt{}122.61=11.07$

If $\small X_1$ and $\small X_2$ are two independent random variables with and , what is the standard deviation of the sum, $\small X_1+X_2?$

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If the random variables are independent, the variances are additive in the sense that

$\small Var(X+Y)=Var(X)+Var(Y)$ .

So then the variance of the sum is

$\small \small Var(X_1+X_2)=Var(X_1)+Var(X_2)=3+3=6$ .

The standard deviation is the square root of the variance, so we have

$\small SD(X_1+X_2)=$\sqrt{6}$$ .

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

$\small X=1$ with $\small P(X=1)=\frac{1}{8}$$

$\small X=2$ with $\small P(X=2)=\frac{3}{8}$$

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

Compute the probability $\small P(X\ge2)$ .

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This probability is simple to compute:

We want to add the probability that X is greater or equal to two. This means the probability that X=2 or X=3.

$\small \small P(X\ge 2)=P(X=2 \or X=3)$

Adding the necessary probabilities we arrive at the solution.

$\small =\frac{3}{8}$+\frac{1}{2}$=\frac{7}{8}$$

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

$\small X=1$ 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 expected value of the distribution.

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The expected value is computed as

$\small \mathbb{E}[X]=\small $\sum_{x}$ xP(X=x)$

for any values of x that the random variable takes.

So we have

$\small \small \mathbb{E}[X]= $\frac{1}{4}$\cdot 1+\frac{1}{4}$\cdot 2+\frac{1}{4}$\cdot 3+\frac{1}{4}$\cdot 4$

$\small =\frac{10}{4}$=2.5$

The average number of calories in a Lick Yo' Lips lollipop is , with a standard deviation of . The calories per lollipop are normally distributed, so what percent of lollipops have more than calories?

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The random variable number of calories per lollipop, so the answer is

$P(\mbox{Number of Calories} > 225)$ or

$P(X > 225) = P(Z > $\frac{(225-210)}{10}$) = P(Z > 1.5) = 1- 0.9332 = 0.0668.$

We have two independent, normally distributed random variables $\small X_1$ and $\small X_2$ such that $\small X_1$ has mean $\small 0$ and variance $\small 2$ and $\small X_2$ has mean $\small 0$ and variance $\small 1$ . What is the probability distribution of the difference of the random variables, $\small \small X_1-X_2$ ?

Tap to see back →

The mean for any set of random variables is additive in the sense that

$\small Mean(X_1+X_2)=Mean(X_1)+Mean(X_2)$

The difference is also additive, so we have

$\small Mean(X_1-X_2)=Mean(X_1)-Mean(X_2)$

This means the mean of $\small X_1-X_2$ is $\small 0$ .

The variance is additive when the random variables are independent, which they are in this case. But it's additive in the sense that for any real numbers $\small a,b$ (even when negative), we have

$\small Var(aX_1+bX_$2)=a^2Var$(X_$1)+b^2Var$(X_2)$ .

So for this difference, we have

$\small \small \small Var(X_1-X_2)=Var(X_1+(-1)X_$2)=1^2Var$(X_$1)+(-1)^2Var$(X_2)$

$\small =Var(X_1)+Var(X_2)=2+1=3$ .

So the mean and variance are $\small 0$ and $\small 3$ , respectively. In addition to that, $\small X_1-X_2$ is normally distributed because the sum or difference of any set of independent normal random variables is also normally distributed.

Robert's work schedule for next week will be released today. Robert will work either 45, 40, 25, or 12 hours. The probabilities for each possibility are listed below:

45 hours: 0.3

40 hours: 0.2

25 hours: 0.4

12 hours: 0.1

What is the standard deviation of the possible outcomes?

Tap to see back →

There are four steps to finding the standard deviation of random variables. First, calculate the mean of the random variables. Second, for each value in the group (45, 40, 25, and 12), subtract the mean from each and multiply the result by the probability of that outcome occurring. Third, add the four results together. Fourth, find the square root of the result.

$\sqrt{}122.61=11.07$

If $\small X_1$ and $\small X_2$ are two independent random variables with and , what is the standard deviation of the sum, $\small X_1+X_2?$

Tap to see back →

If the random variables are independent, the variances are additive in the sense that

$\small Var(X+Y)=Var(X)+Var(Y)$ .

So then the variance of the sum is

$\small \small Var(X_1+X_2)=Var(X_1)+Var(X_2)=3+3=6$ .

The standard deviation is the square root of the variance, so we have

$\small SD(X_1+X_2)=$\sqrt{6}$$ .

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

$\small X=1$ with $\small P(X=1)=\frac{1}{8}$$

$\small X=2$ with $\small P(X=2)=\frac{3}{8}$$

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

Compute the probability $\small P(X\ge2)$ .

Tap to see back →

This probability is simple to compute:

We want to add the probability that X is greater or equal to two. This means the probability that X=2 or X=3.

$\small \small P(X\ge 2)=P(X=2 \or X=3)$

Adding the necessary probabilities we arrive at the solution.

$\small =\frac{3}{8}$+\frac{1}{2}$=\frac{7}{8}$$

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

$\small X=1$ 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 expected value of the distribution.

Tap to see back →

The expected value is computed as

$\small \mathbb{E}[X]=\small $\sum_{x}$ xP(X=x)$

for any values of x that the random variable takes.

So we have

$\small \small \mathbb{E}[X]= $\frac{1}{4}$\cdot 1+\frac{1}{4}$\cdot 2+\frac{1}{4}$\cdot 3+\frac{1}{4}$\cdot 4$

$\small =\frac{10}{4}$=2.5$

The average number of calories in a Lick Yo' Lips lollipop is , with a standard deviation of . The calories per lollipop are normally distributed, so what percent of lollipops have more than calories?

Tap to see back →

The random variable number of calories per lollipop, so the answer is

$P(\mbox{Number of Calories} > 225)$ or

$P(X > 225) = P(Z > $\frac{(225-210)}{10}$) = P(Z > 1.5) = 1- 0.9332 = 0.0668.$

Robert's work schedule for next week will be released today. Robert will work either 45, 40, 25, or 12 hours. The probabilities for each possibility are listed below:

45 hours: 0.3

40 hours: 0.2

25 hours: 0.4

12 hours: 0.1

What is the standard deviation of the possible outcomes?

Tap to see back →

There are four steps to finding the standard deviation of random variables. First, calculate the mean of the random variables. Second, for each value in the group (45, 40, 25, and 12), subtract the mean from each and multiply the result by the probability of that outcome occurring. Third, add the four results together. Fourth, find the square root of the result.

$\sqrt{}122.61=11.07$

We have two independent, normally distributed random variables $\small X_1$ and $\small X_2$ such that $\small X_1$ has mean $\small 0$ and variance $\small 2$ and $\small X_2$ has mean $\small 0$ and variance $\small 1$ . What is the probability distribution of the difference of the random variables, $\small \small X_1-X_2$ ?

Tap to see back →

The mean for any set of random variables is additive in the sense that

$\small Mean(X_1+X_2)=Mean(X_1)+Mean(X_2)$

The difference is also additive, so we have

$\small Mean(X_1-X_2)=Mean(X_1)-Mean(X_2)$

This means the mean of $\small X_1-X_2$ is $\small 0$ .

The variance is additive when the random variables are independent, which they are in this case. But it's additive in the sense that for any real numbers $\small a,b$ (even when negative), we have

$\small Var(aX_1+bX_$2)=a^2Var$(X_$1)+b^2Var$(X_2)$ .

So for this difference, we have

$\small \small \small Var(X_1-X_2)=Var(X_1+(-1)X_$2)=1^2Var$(X_$1)+(-1)^2Var$(X_2)$

$\small =Var(X_1)+Var(X_2)=2+1=3$ .

So the mean and variance are $\small 0$ and $\small 3$ , respectively. In addition to that, $\small X_1-X_2$ is normally distributed because the sum or difference of any set of independent normal random variables is also normally distributed.

If $\small X_1$ and $\small X_2$ are two independent random variables with and , what is the standard deviation of the sum, $\small X_1+X_2?$

Tap to see back →

If the random variables are independent, the variances are additive in the sense that

$\small Var(X+Y)=Var(X)+Var(Y)$ .

So then the variance of the sum is

$\small \small Var(X_1+X_2)=Var(X_1)+Var(X_2)=3+3=6$ .

The standard deviation is the square root of the variance, so we have

$\small SD(X_1+X_2)=$\sqrt{6}$$ .

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

$\small X=1$ with $\small P(X=1)=\frac{1}{8}$$

$\small X=2$ with $\small P(X=2)=\frac{3}{8}$$

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

Compute the probability $\small P(X\ge2)$ .

Tap to see back →

This probability is simple to compute:

We want to add the probability that X is greater or equal to two. This means the probability that X=2 or X=3.

$\small \small P(X\ge 2)=P(X=2 \or X=3)$

Adding the necessary probabilities we arrive at the solution.

$\small =\frac{3}{8}$+\frac{1}{2}$=\frac{7}{8}$$

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

$\small X=1$ 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 expected value of the distribution.

Tap to see back →

The expected value is computed as

$\small \mathbb{E}[X]=\small $\sum_{x}$ xP(X=x)$

for any values of x that the random variable takes.

So we have

$\small \small \mathbb{E}[X]= $\frac{1}{4}$\cdot 1+\frac{1}{4}$\cdot 2+\frac{1}{4}$\cdot 3+\frac{1}{4}$\cdot 4$

$\small =\frac{10}{4}$=2.5$

The average number of calories in a Lick Yo' Lips lollipop is , with a standard deviation of . The calories per lollipop are normally distributed, so what percent of lollipops have more than calories?

Tap to see back →

The random variable number of calories per lollipop, so the answer is

$P(\mbox{Number of Calories} > 225)$ or

$P(X > 225) = P(Z > $\frac{(225-210)}{10}$) = P(Z > 1.5) = 1- 0.9332 = 0.0668.$

Robert's work schedule for next week will be released today. Robert will work either 45, 40, 25, or 12 hours. The probabilities for each possibility are listed below:

45 hours: 0.3

40 hours: 0.2

25 hours: 0.4

12 hours: 0.1

What is the standard deviation of the possible outcomes?

Tap to see back →

There are four steps to finding the standard deviation of random variables. First, calculate the mean of the random variables. Second, for each value in the group (45, 40, 25, and 12), subtract the mean from each and multiply the result by the probability of that outcome occurring. Third, add the four results together. Fourth, find the square root of the result.

$\sqrt{}122.61=11.07$

We have two independent, normally distributed random variables $\small X_1$ and $\small X_2$ such that $\small X_1$ has mean $\small 0$ and variance $\small 2$ and $\small X_2$ has mean $\small 0$ and variance $\small 1$ . What is the probability distribution of the difference of the random variables, $\small \small X_1-X_2$ ?

Tap to see back →

The mean for any set of random variables is additive in the sense that

$\small Mean(X_1+X_2)=Mean(X_1)+Mean(X_2)$

The difference is also additive, so we have

$\small Mean(X_1-X_2)=Mean(X_1)-Mean(X_2)$

This means the mean of $\small X_1-X_2$ is $\small 0$ .

The variance is additive when the random variables are independent, which they are in this case. But it's additive in the sense that for any real numbers $\small a,b$ (even when negative), we have

$\small Var(aX_1+bX_$2)=a^2Var$(X_$1)+b^2Var$(X_2)$ .

So for this difference, we have

$\small \small \small Var(X_1-X_2)=Var(X_1+(-1)X_$2)=1^2Var$(X_$1)+(-1)^2Var$(X_2)$

$\small =Var(X_1)+Var(X_2)=2+1=3$ .

So the mean and variance are $\small 0$ and $\small 3$ , respectively. In addition to that, $\small X_1-X_2$ is normally distributed because the sum or difference of any set of independent normal random variables is also normally distributed.