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GRE Subject Test: Chemistry : Variance and Standard Deviation

Study concepts, example questions & explanations for GRE Subject Test: Chemistry

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Example Questions

Example Question #5 : How To Find Standard Deviation Of A Random Variable

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?

Possible Answers:

32.7

122.61

13.5

11.07

35.8

Correct answer:

11.07

Explanation:

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.

$(45-32.7)^{2}(0.3)=45.38$

$(40-32.7)^{2}(0.2)=10.66$

$(25-32.7)^{2}(0.4)=23.72$

$(12-32.7)^{2}(0.1)=42.85$

45.38+10.66+23.72+42.85=122.61

$\sqrt{}122.61=11.07$

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Example Question #1 : How To Find Standard Deviation Of A Random Variable

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, $\small \small X_1-X_2$ ?

Possible Answers:

Normal distribution with mean $\small 1$ and variance $\small 3$ .

Normal distribution with mean $\small 1$ and variance $\small 2$ .

Normal distribution with mean $\small 0$ and variance $\small 3$ .

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

Correct answer:

Normal distribution with mean $\small 0$ and variance $\small 3$ .

Explanation:

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.

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GRE Subject Test: Chemistry : Variance and Standard Deviation

Study concepts, example questions & explanations for GRE Subject Test: Chemistry

All GRE Subject Test: Chemistry Resources

Example Questions

Example Question #5 : How To Find Standard Deviation Of A Random Variable

Example Question #1 : How To Find Standard Deviation Of A Random Variable

All GRE Subject Test: Chemistry Resources

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