Question 1

The proportion of the total area under a normal curve between two z-scores corresponds to the ______ of that range of scores.

Accepted Answer

The proportion of the total area under a normal curve between two z-scores is a relative frequency. This is because it represents the proportion of scores within that range relative to the total number of scores. The other choices are not accurate because numeric magnitude refers to the value of individual scores, variability refers to the spread of scores, and frequency refers to the number of times a score occurs in a distribution.

Question 2

The average score on the statistics quiz was a 70,and there was a standard deviation of 10 points.If Enrique had a z-score of -1.0,his raw score would be ______ and his percentile rank would be ______.

Accepted Answer

To find the raw score, we use the formula:

z-score = (raw score - mean) / standard deviation

Substituting the given values, we get:

1.0 = (raw score - 70) / 10

Solving for the raw score, we get:

raw score = 70 + 10(1.0) = 80

To find the percentile rank, we use a z-table and look for the area to the left of the z-score. From the table, we find that the area to the left of a z-score of 1.0 is 0.8413, or 84.13%.

However, we need to find the percentile rank below the score, so we subtract the area from 1 and convert to a percentage:

percentile rank = (1 - 0.8413) x 100% ≈ 15.87%

Therefore, the best choice is C) 60; 16%.

Question 3

When two normal z-distributions are plotted on the same graph,what can we say about the relative frequency of each z-score?

Accepted Answer

Since z-scores standardize different distributions for comparison by accounting for the mean and standard deviation of each distribution, two normal z-distributions plotted on the same graph will have the same relative frequency for each z-score. This is because z-scores are a way to describe the position of a raw score within a distribution in terms of how many standard deviations away from the mean it is.

Question 4

If there were 500 students in Jamal's class,approximately how many actual students scored higher than Jamal on the quiz if Jamal had a z-score of -1?

Accepted Answer

To find out how many students scored higher than Jamal, we need to convert his z-score to a percentile. Using a standard normal distribution table, we can see that a z-score of 1.30 corresponds to a percentile of approximately 90.31%. This means that Jamal scored higher than approximately 90.31% of the class, leaving approximately 100% - 90.31% = 9.69% of the class who scored higher than him. To find the actual number of students, we can multiply 9.69% by the total number of students: 0.0969 x 500 = 48.45. Since we can't have a fractional part of a student, we round up to the nearest whole number, giving us an answer of approximately 49. However, we need to remember that this is an estimate, so the closest answer choice is C, which gives an estimate of 420 students scoring higher than Jamal.

Question 5

A perfect normal curve,which serves as a model of any approximately normal z-distribution,is called the

Accepted Answer

The perfect normal curve that serves as a model for any approximately normal z-distribution is called the standard normal curve or the standard normal distribution. It has a mean of 0 and a standard deviation of 1, and all scores are represented as z-scores. Therefore, the correct answer is C, the standard normal curve. Choices A, B, and D are not correct because they either refer to different types of distributions or do not exist as standard statistical terms.

Question 6

If only the upper 30% of a normally distributed class passed a quiz for which the mean was 70 and the standard deviation was 10,what was the lowest score a student could have received and still have passed?

Accepted Answer

First, we need to find the z-score associated with the upper 30% of scores. To do this, we can use a z-table or calculator and find the z-score associated with a cumulative probability of 0.7 (since we want the upper 30%, which is the same as the lower 70%). We get a z-score of approximately 0.524.

Next, we can use the formula z = (x - μ) / σ to solve for x, the lowest passing score. Plugging in the values we know: 
0.524 = (x - 70) / 10

Solving for x, we get: 
x = (0.524)(10) + 70 = 75.24

Therefore, the lowest passing score is 75.24, which rounds to 75.2. So, the answer is (D) 75.2.

Question 7

Your sense of direction puts you at the 39th percentile,relative to a raw score distribution with a mean of 3.5 and a standard deviation of 1.What is your z-score?

Accepted Answer

The z-score is calculated as (x - mu) / sigma, where x is the raw score, mu is the mean, and sigma is the standard deviation. In this case, x = 39th percentile, mu = 3.5, and sigma = 1. So, the z-score is (39 - 3.5) / 1 = 35.5 / 1 = 35.5. The correct answer is A.

Question 8

Only the upper 30% of a normally distributed class passed a quiz for which the mean was 70 and the standard deviation was 10.In this class,76.7 and 78.4 are the boundary limits for students who scored between what percentiles?

Accepted Answer

The z-score for the 30th percentile is -0.524. The z-score for the 70th percentile is 0.524. The z-score for 76.7 is (76.7 - 70) / 10 = 0.67. The z-score for 78.4 is (78.4 - 70) / 10 = 0.84. The percentiles corresponding to these z-scores are 75th and 80th.

Question 9

Which of the following is NOT a way we apply the standard normal curve?

Accepted Answer

The answer of Which of the following is NOT a...

Question 10

The mean of a z-score distribution is always ______,and the standard deviation is always ______.

Accepted Answer

The mean of a z-score distribution is always 0 because the formula for calculating z-scores involves subtracting the mean from the raw score and dividing by the standard deviation, resulting in a distribution with a mean of 0. The standard deviation of a z-score distribution is always 1 because the formula for calculating z-scores involves dividing by the standard deviation, resulting in a distribution with a standard deviation of 1. Therefore, the correct answer is D, with a mean of 0 and a standard deviation of 1.

Question 11

In a z-distribution,the standard deviation will always be

Accepted Answer

In a z-distribution, the standard deviation is always equal to 1. This is because z-scores are calculated by subtracting the mean and dividing by the standard deviation of the distribution. Since the mean of a z-distribution is always 0 and the standard deviation is always 1, z-scores allow us to compare different normal distributions using a standardized scale. Thus, option C is the correct answer.

Question 12

The distribution of z-scores is always

Accepted Answer

The distribution of z-scores is always the same as the distribution of raw scores, but with a mean of 0 and a standard deviation of 1. It is not necessarily positively or negatively skewed, and may be more or less spread out than the distribution of raw scores depending on the shape and variability of the original distribution.

Question 13

How accurately the standard normal curve model predicts the actual relative frequency of raw scores depends on three aspects of the data:

Accepted Answer

The standard normal curve model predicts the actual relative frequency of raw scores accurately when the distribution of these scores is approximately normal, the sample size is large, and the scores are continuous and measured on an interval or ratio scale. These conditions ensure the applicability of the normal distribution model.

Question 14

If a class's scores are normally distributed,with a mean of 70 and a standard deviation of 10,what are the upper and lower limits of the middle 68% of the class?

Accepted Answer

The middle 68% of a normally distributed dataset falls within one standard deviation from the mean. Since the mean is 70 and the standard deviation is 10, one standard deviation below the mean is 60 (70-10) and one standard deviation above the mean is 80 (70+10). Therefore, the upper and lower limits of the middle 68% of the class are 60 and 80.

Question 15

Your sense of direction puts you at the 79th percentile,relative to a raw score distribution with a mean of 3.5 and a standard deviation of 1.What is your z-score?

Accepted Answer

To find the z-score, we need to use the formula:
z = (raw score - mean) / standard deviation
Given that the raw score is at the 79th percentile, we know that 79% of the scores are below this score. Therefore, the area to the left of the score on the normal distribution is 0.79. Using a standard normal distribution table, we can find the z-score corresponding to this area as approximately 0.81.
z = (x - 3.5) / 1, where x is the raw score
0.81 = (x - 3.5) / 1
x - 3.5 = 0.81
x = 4.31
Therefore, the z-score is (4.31 - 3.5) / 1 = 0.81.

Question 16

If raw scores ranging from 1 to 50 represented all the corresponding negative z-scores on the distribution,and raw scores ranging from 50 to 100 represented all the corresponding positive z-scores,what would be the total relative frequency of 50 to 100?

Accepted Answer

Since all the raw scores between 50 to 100 represent positive z-scores, we can calculate the area under the standard normal curve from z = 0 (which corresponds to a raw score of 50) to z = infinity (which corresponds to a raw score of 100). Using a standard normal table or calculator, we find that this area is approximately 0.1915. Since the total area under the standard normal curve is 1, the relative frequency of raw scores between 50 to 100 is:

Relative frequency = area between 50 to 100 / total area = 0.1915 / 1 = 0.50

Therefore, the best choice is C.

Question 17

On a recent paper,Maria's score,transformed to a z-score,was +1.If the cutoff mark is the upper 10% of the class for As,the upper 10% to 20% for Bs,the upper 20% to 30% for Cs,and the upper 30% to 40% for Ds (below 60% being a failure),what grade did Maria receive on the quiz?

Accepted Answer

A z-score of +1 corresponds to being one standard deviation above the mean. The upper 10% of the class would be within approximately 1.28 standard deviations above the mean. Therefore, Maria's score falls within the range for a B.

Question 18

To evaluate a person for possible brain damage,a neuropsychologist gives the person a visual memory test and a reading test.To compare the person's performance across these two tests,what should the neuropsychologist do?

Accepted Answer

The best choice is to calculate a z-score for each test because this allows for comparison of performance on two different tests with potentially different scoring scales. By standardizing each person's score based on the mean and standard deviation of their respective sample population, it becomes easier to compare their performance relative to their peers. Graphing the raw score distribution on the same graph may be useful in visualizing the distribution of scores, but it does not allow for direct comparison of performance on the two different tests. Calculating z-scores for the sample means would not be relevant for evaluating an individual's performance, and finding the simple frequency of the person's raw score does not provide a way to compare their performance relative to the larger population.

Question 19

If your sense of direction,relative to a sample of college students,gave you a z-score of +1.5,what is your percentile?

Accepted Answer

A z-score of +1.5 indicates that the person is at the 93rd percentile, meaning they score higher than 93% of the sample of college students.

Question 20

We can use the standard normal curve as our model for

Accepted Answer

The standard normal curve can be used as a model for any approximately normal distribution when transformed to z-scores. This is because the process of standardizing a distribution by transforming it into z-scores converts any distribution into a standard normal distribution with a mean of 0 and a standard deviation of 1. Therefore, using the standard normal curve can help us make statistical inferences and calculations based on the properties of the standard normal distribution, regardless of the specific shape of the original distribution.

Quiz 6: Z-Scores and the Normal Curve