GRE Subject Test: Math : Probability & Statistics

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

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Example Questions

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Example Question #1 : How To Find Z Scores For A Data Set

Natalie took her university placement examinations in Spanish and math. In Spanish, she scored 82; in math, she scored 86. The results of the Spanish exam had a mean of 72 and a standard deviation of 8. The results of the math exam had a mean of 68 and a standard deviation of 12. On which exam did Natalie do better, compared with the rest of her peers taking these placement exams?

Possible Answers:

The Spanish exam

The Math exam

She did equally well on both exams

None of the other answers, as z-scores cannot be calculated for this question

Correct answer:

The Math exam

Explanation:

We need to calculate Natalie's z-scores for both her Spanish and math exams. Calculating z-scores is as follows:

Her z-score for the Spanish exam is , which equals 1.25, while her z-score for the math exam is , which equals 1.50. Since both z-scores are positive, Natalie did above average on both tests, but since her z-score for the math exam is higher than her z-score for the Spanish exam, Natalie did better on her math exam when compared to the rest of her peers taking the exam.

Example Question #1 : How To Find Z Scores For A Data Set

This year's harvest, the apples had a mean mass of  with a standard deviation of . You pick an apple from the harvest and find its mass to be .

What is the z-score of the apple you picked?

Possible Answers:

Correct answer:

Explanation:

To find the z-score, we follow the formula

 where  is the given value,  is the mean, and  is the standard deviation.

For this problem we see that

    and 

Substituting for these values we see

Example Question #1 : Z Distribution

A bread maker has an average life of 9 years, with a variance of 4 years. What is the z-score associated with the probability that a bread maker lasts less than 13 years?

Possible Answers:

Correct answer:

Explanation:

 is the average life of the bread maker, in this problem 9 years.

 is the standard deviation of the break makers life. The standard deviation is the square root of the variance. 

X is the test statistic we are looking to find. In this problem it is the probability that the bread maker lasts less than 13 years.

If you were to find the value associated with a z-score of 2 on the z-table, you would see that

Example Question #111 : Probability & Statistics

The purpose of the t test is to do which of the following?

Possible Answers:

Compare the means of separate populations.

Describe the variance within a population.

Compare the significance of multiple variables.

Determine the ability of one variable to predict another.

Test the relationship between five or more variables.

Correct answer:

Compare the means of separate populations.

Explanation:

A t test is used to compare the means of different groups.  A t test score describes the likelihood that the difference in means between two groups is due to chance. The null hypothesis assumes the two sets are equal however, one can reject the null hypothesis with a p value within a particular confidence level.

Example Question #181 : Ap Statistics

follows a chi-squared distribution with degrees of freedom and through are independent standard normal variables.

If , what is ?

Possible Answers:

Correct answer:

Explanation:

Use the fact that the sum of  squared standard and independent normal variables follows a chi-squared distribution with degrees of freedom.

Example Question #1 : Confidence Intervals & Hypothesis Testing

A sample of  observations of 02 consumption by adult western fence lizards gave the following statistics:

Find the  confidence limit for the mean 02 consumption by adult western fence lizards.

Possible Answers:

Correct answer:

Explanation:

Because we are only given the sample standard deviation we will use the t-distribution to calculate the confidence interval.

Appropriate Formula: 

Now we must identify our variables:

We must find the appropriate t-value based on the given

t-value at 90% confidence:

 

Look up t-value for 0.05, 55 , so t-value= ~ 1.6735

90% CI becomes:

 

Example Question #1 : Confidence Intervals And Mean

Subject     

Horn Length (in)

Subject      

Horn Length (in)

1

19.1

11

11.6

2

14.7

12

18.5

3

10.2

13

28.7

4

16.1

14

15.3

5

13.9

15

13.5

6

12.0

16

7.7

7

20.7

17

17.2

8

8.6

18

19.0

9

24.2

19

20.9

10

17.3

20

21.3

The data above represents measurements of the horn lengths of African water buffalo that were raised on calcium supplements. Construct a 95% confidence interval for the population mean for horn length after supplments.

Possible Answers:

Correct answer:

Explanation:

First you must calculate the sample mean and sample standard deviation of the sample.

Because we do not know the population standard deviation we will use the t-distribution to calculate the confidence intervals. We must use standard error in this formula because we are working with the standard deviation of the sampling distribution.

Formula: 

To find the appropriate t-value for 95% confidence interval: 

Look up in t-table and the corresponding t-value = 2.093.

Thus the 95% confidence interval is: 

 

Example Question #1 : Confidence Intervals And Mean

The 95% confidence interval created for the difference in means between two training programs for middle distance college runners is . The variable being measured is the improvement in seconds of mile times over the course of a season. One program has more speed work and intervals, while the other focuses more on distance training.

What does the confidence interval tell us about the difference in the two programs?

Possible Answers:

The mean improvement of  second  is too small to matter, so reject the null. There is no evidence that one program is better at reducing miles times.

The confidence interval is large, so one program is clearly better at reducing mile times than the other.

Zero is in the interval, so do not reject the null. There is no evidence that one program is better than the other.

 is greater than , so reject the null. This is evidence that one program is better at reducing mile times than the other.

Zero is not in the interval, so reject the null. This is evidence that one program is significantly better at reducing mile times.

Correct answer:

Zero is in the interval, so do not reject the null. There is no evidence that one program is better than the other.

Explanation:

For there to be a statistically significant difference in the training programs, the 95% confidence interval cannot include zero.  includes zero, so we can't say that one program is significantly better than the other.

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