Question 1

In a sample of n = 40 scores,X = 31 corresponds to z = -1.50 and X = 39 corresponds to z = +0.50.The sample mean is M = 36.

Accepted Answer

The sample mean is not equivalent to the mean of the standardized values, which is always 0. Therefore, X = 31 corresponding to z = -1.50 and X = 39 corresponding to z = +0.50 have no bearing on the value of the sample mean, M = 36.

Question 2

If an entire population of N = 20 scores is transformed into z-scores,the set of 20 z-scores will have X = 0.

Accepted Answer

By definition, the sum of all z-scores in a population is always equal to 0.

Question 3

If an entire population of N = 20 scores is transformed into z-scores,the set of 20 z-scores will have SS = 20.

Accepted Answer

The answer of If an entire population of N =...

Question 4

On an exam,Tom scored 12 points above the mean and had a z-score of +2.00.The standard deviation for the set of exam scores must be σ = 6.

Accepted Answer

Since Tom's z-score is +2.00, which means he scored 2 standard deviations above the mean. We know that he scored 12 points above the mean, which means the distance between Tom's score and the mean is 2 standard deviations or 2 x 6 = 12 points. Therefore, the mean is 12 points below Tom's score, which is equivalent to Tom's score being 12 points above the mean. Hence, the statement is true, and the answer is A.

Question 5

For a sample with a mean of M = 76,a score of X = 72 corresponds to z = -0.50.The sample standard deviation is s = 8.

Accepted Answer

We can use the formula for z-score:

z = (X - M) / s

Plugging in the values we know, we get:

-0.5 = (72 - 76) / 8

Solving for M, we get:

M = 76 + (-0.5) * 8

M = 72

Since the given score of X = 72 corresponds to z = -0.5, this calculation confirms the mean of the sample as M = 76. Therefore, the statement is true.

Question 6

For a sample with M = 50 and s = 8,z = 1.50 corresponds to X = 65.

Accepted Answer

The formula to find a score (X) given a z-score is X = M + z(s), where M is the mean and s is the standard deviation. Plugging in the values given: X = 50 + 1.50(8) = 50 + 12 = 62, not 65.

Question 7

For a population with μ = 34,a score of X = 31 corresponds to z = -1.00.The standard deviation for the population is σ = 6.

Accepted Answer

The z-score is calculated using the formula $z = \frac{(X - \mu)}{\sigma}$. Plugging in the given values, $z = \frac{(31 - 34)}{6} = \frac{-3}{6} = -0.5$, not -1.00.

Question 8

For a sample with a standard deviation of s = 10,a score with a deviation of +5 will have a z-score of z = 0.50.

Accepted Answer

This is true because z-score is calculated as the difference between the score and the mean divided by the standard deviation. In this case, the score is 5 above the mean (deviation of +5) and the standard deviation is 10. Therefore, the z-score = (5-0)/10 = 0.50.

Question 9

Transforming an entire distribution of scores into z-scores will not change the shape of the distribution.

Accepted Answer

Transforming scores into z-scores standardizes the distribution such that the mean is 0 and the standard deviation is 1. This does not change the shape of the distribution, but only the scale.

Question 10

A population with μ = 54 and σ = 12 is standardized to create a new distribution with μ = 100 and σ = 20.After the transformation,an individual receives a new score of X = 95.The original score for this individual was X = 51.

Accepted Answer

To change a score from the original distribution to the standardized distribution, we use the formula z = (X - μ) / σ. In this case, z = (51 - 54) / 12 = -0.25. 
To change a score from the standardized distribution to the original distribution, we use the formula X = μ + (z * σ). In this case, X = 100 + (-0.25 * 20) = 95. 
Therefore, it is true that the individual received a new score of 95 in the standardized distribution and their original score was 51 in the original distribution.

Question 11

In a population with σ = 4,a score of X = 48 corresponds to z = 1.50.The mean for this population is µ = 42.

Accepted Answer

Since we know the population standard deviation (σ), we can use the formula z = (X - µ) / σ to find the z-score for a given score. Plugging in X = 48, σ = 4, and z = 1.50, we get:

1.5 = (48 - 42) / 4

Solving for the mean (µ), we get:

µ = X - zσ
µ = 48 - 1.5(4)
µ = 42

So the given mean (µ = 42) is consistent with the given score (X = 48) and z-score (z = 1.50), and therefore the statement is true.

Question 12

For a sample with a mean of M = 50 and a standard deviation of s = 10,a z-score of
z = +2.00 corresponds to X = 70.

Accepted Answer

This is true because the formula for a z-score is (X - M) / s. If we plug in the values given, we get (70 - 50) / 10 = 2, which corresponds to the z-score of +2.00.

Question 13

A professor standardizes exam scores so that all exams have μ = 50 and σ = 10.If the original scores from an exam have μ = 42 and σ = 6,then a student with an original exam score of X = 45 would receive a standardized score of X = 55.

Accepted Answer

Since the original mean and standard deviation are lower than the standardized mean and standard deviation, a score of 45 would actually be below average on the original scale but above average on the standardized scale, resulting in a higher standardized score of 55.

Question 14

For a sample with a standard deviation of s = 8,a score of X = 42 corresponds to
z = -0.25.The mean for the sample is M = 40.

Accepted Answer

The answer of For a sample with a standard deviation...

Question 15

For a population of exam scores,a score of X = 83 corresponds to z = +0.50 and a score of X = 89 corresponds to z = +1.50.The population mean is μ = 80.

Accepted Answer

This is true because we can use the formula z = (X - μ) / σ, where μ is the population mean, X is the individual score, and σ is the population standard deviation. Using the given information, we can set up the equations:

0.5 = (83 - 80) / σ
1.5 = (89 - 80) / σ

Solving for σ in both equations, we get σ = 6 and σ = 3.33, respectively. Since we cannot have two different standard deviations for the same population, we can use the average of these two values, which is (6 + 3.33) / 2 = 4.67. Now, we can use the first equation to find the population standard deviation:

0.5 = (83 - 80) / 4.67
4.67(0.5) = 83 - 80
2.335 = 3

Therefore, the population standard deviation is σ = 3, and we can check if the second equation holds true:

1.5 = (89 - 80) / 3
4.5 = 9

This is true, so we can conclude that a score of X = 83 has a z-score of +0.50 and a score of X = 89 has a z-score of +1.50, given a population mean of μ = 80.

Question 16

On an exam with σ = 6,Tom's score of X = 54 corresponds to z = -1.00.The mean for the exam must be µ = 60.

Accepted Answer

The formula for calculating a z-score is z = (X - µ) / σ, where X is the score, µ is the mean, and σ is the standard deviation. Plugging in the given values: -1 = (54 - µ) / 6. Solving for µ gives µ = 60.

Question 17

For a sample with a standard deviation of s = 6,a z-score of z = -1.50 corresponds to a location that is 9 points above the mean.

Accepted Answer

A z-score of -1.50 corresponds to a location that is 1.5 standard deviations below the mean, not above it. Therefore, the correct statement would be that the location is 9 points below the mean (since each standard deviation is equal to 6).

Question 18

For a population with a mean of μ = 40,a score of X = 37 corresponds to z = -0.50.The standard deviation for the population is σ = 3.

Accepted Answer

The answer of For a population with a mean of...

Question 19

In a sample with s = 6,a score of X = 53 corresponds to z = -0.50.The mean for this sample is M = 56.

Accepted Answer

This is true because we can use the formula z = (X - M)/s to solve for the value of s. Plugging in the given values, we get: -0.50 = (53 - 56)/s. Solving for s, we get s = 6. Therefore, we know that the standard deviation of the sample is 6, and we also know that the mean is M = 56.

Question 20

A population with μ = 41 and σ = 4 is standardized to create a new distribution with μ =
50 and σ = 10.In this transformation,a score of X = 39 from the original distribution will be
transformed into a score of X = 55.

Accepted Answer

It is not possible for a score of X = 39 from the original distribution to be transformed into a score of X = 55 in the standardized distribution, as this would require a z-score of approximately 3.5, which is beyond the range of possible z-scores. Therefore, the statement is false.

Quiz 5: Z-Scores