Question 1

Using a normal curve table,the Z score a person would need to be in the top 4% of his or her class on a particular test is
A)4% − 50% = -46%,which corresponds to a Z score of -1.75
B)50% − 46% = 4%,which corresponds to a Z score of 0.10
C)50% − 4% = 46%,which corresponds to a Z score of 1.75
D)34% + 14% − 4% = 44%,which corresponds to a Z score of 1.55 or 1.56

Accepted Answer

To be in the top 4%, the person needs to be above 96% of the class. Using the normal curve table, the Z score for the 96th percentile is approximately 1.75.

Question 2

Using a normal curve table,if a person has a score in the bottom 42% of all scores in a distribution,the highest Z score this person can have is
A)-1.41
B)-0.94
C)-0.20
D)0)25

Accepted Answer

C

Question 3

Using a normal curve table,if a person received a test score that is in the top 32% of all test scores,the person's Z score must be at least
A)0.46
B)7.14
C)12.55
D)0.91

Accepted Answer

The area to the left of a Z score representing the top 32% is equal to 1 - 0.32 = 0.68. Using a normal curve table, we can find the Z score that corresponds to an area of 0.68 to be approximately 0.46. Therefore, if a person received a test score that is in the top 32% of all test scores, their Z score must be at least 0.46.

Question 4

In a distribution of Z scores,the mean is always
A)10
B)1
C)0
D)50

Accepted Answer

In a distribution of Z scores, the mean is always 0 because it represents the standard deviation from the mean of the original distribution.

Question 5

Molly wants to pursue a graduate degree at New University but is unsure whether to specialize in law or medicine. To help her decide, she takes the MCAT and LSAT exams.
LSAT: Molly's score = 120; Students at New University: M = 150, SD = 15
MCAT: Molly's score = 52; Students at New University: M = 40, SD = 6
-Molly's Z scores were:
A)LSAT,Z = 2.00; MCAT Z = 2.00
B)LSAT,Z = 2.00; MCAT Z = -2.00
C)LSAT,Z = -2.00; MCAT Z = 2.00
D)LSAT,Z = -2.00; MCAT Z = -2.00

Accepted Answer

LSAT Z score = (120 - 150) / 15 = -2.00; MCAT Z score = (52 - 40) / 6 = 2.00.

Question 6

If the mean score on a creativity test is 20,the standard deviation is 5,and the distribution is normal,the percentage of people who would obtain scores between 15 and 25 is
A)34%
B)14% + 34% = 48%
C)50%
D)34% + 34% = 68%

Accepted Answer

We can use the empirical rule to estimate the percentage of people who would obtain scores between 15 and 25. According to the empirical rule, approximately 68% of the scores fall within one standard deviation of the mean in a normal distribution.

So, the percentage of people who would obtain scores between 15 and 25 is approximately 34% on each side of the mean (i.e., 20 - 5 = 15 and 20 + 5 = 25).

Therefore, the best choice is D, which is 34% + 34% = 68%.

Question 7

The normal curve is most likely to be described as
A)bimodal and roughly symmetrical.
B)unimodal,roughly symmetrical,and bell-shaped.
C)skewed to the right.
D)a roughly symmetrical U-shape.

Accepted Answer

The questions below (# 8 - 10)are based on the following scenario. Molly wants to pursue a graduate degree at New University but is unsure whether to specialize in law or medicine.To help her decide,she takes the MCAT and LSAT exams. LSAT: Molly's score = 120; Students at New University: M = 150,SD = 15 MCAT: Molly's score = 52; Students at New University: M = 40,SD = 6

Question 8

Molly wants to pursue a graduate degree at New University but is unsure whether to specialize in law or medicine. To help her decide, she takes the MCAT and LSAT exams.
LSAT: Molly's score = 120; Students at New University: M = 150, SD = 15
MCAT: Molly's score = 52; Students at New University: M = 40, SD = 6
-In relation to the other people at New University,on which test did Molly get the better score?
A)the LSAT
B)the MCAT
C)the scores were equal
D)it is impossible to say without more information

Accepted Answer

Molly's performance relative to the other students at New University is better on the MCAT because her score is higher above the mean in terms of standard deviations. For the LSAT, her score is 2 standard deviations below the mean (150-120=30; 30/15=2). For the MCAT, her score is 2 standard deviations above the mean (52-40=12; 12/6=2). Therefore, in terms of standard deviations from the mean, she did better on the MCAT.

Question 9

A psychologist is most likely to calculate Z scores in order to
A)compare scores obtained from two different scales.
B)compare the skew of two different distributions.
C)determine how far above or below the median a score is.
D)determine the shape of a distribution of scores.

Accepted Answer

Z scores are used to standardize scores from different scales so that they can be compared. Z scores reflect how many standard deviations a score is away from the mean of its distribution. This allows for comparisons to be made between different distributions with different means and standard deviations. Therefore, the psychologist is most likely to calculate Z scores to compare scores obtained from two different scales.

Question 10

If the mean of a test is 13 and its standard deviation is 3,the Z score for a person with a raw score of 7 is
A)-2
B)-1
C)1
D)2

Accepted Answer

The Z score is calculated by subtracting the mean from the raw score and then dividing by the standard deviation. For a raw score of 7, the calculation is (7 - 13) / 3 = -6 / 3 = -2.

Question 11

Using the percentage approximations for the normal curve,the percentage of scores between the mean and one standard deviation below the mean is
A)14%
B)50%
C)34%
D)14% + 34% = 48%

Accepted Answer

According to the empirical rule, approximately 68% of scores fall within one standard deviation of the mean. Since the normal curve is symmetric, half of that percentage, or 34%, falls between the mean and one standard deviation below the mean.

Question 12

Molly wants to pursue a graduate degree at New University but is unsure whether to specialize in law or medicine. To help her decide, she takes the MCAT and LSAT exams.
LSAT: Molly's score = 120; Students at New University: M = 150, SD = 15
MCAT: Molly's score = 52; Students at New University: M = 40, SD = 6
-Using the percentage approximations for the normal curve,the percentage of people having a lower score than Molly on the LSAT would be:
A)2%
B)14%
C)48%
D)98%

Accepted Answer

Molly's LSAT score is 120, which is 2 standard deviations below the mean (150 - 2*15 = 120). Approximately 2% of people score lower than 2 standard deviations below the mean in a normal distribution.

Question 13

The variance of a distribution of Z scores is always
A)ΣZ / N
B)(Z)(M + SD)
C)0
D)1

Accepted Answer

The variance of a distribution of Z scores is always 1 because the variance of any standardized distribution is always equal to 1.

Question 14

The raw score that corresponds to a Z score of 2.0 obtained from a distribution with a mean of 80 and a standard deviation of 10 is
A)82
B)90
C)100
D)170

Accepted Answer

The formula to convert a raw score to a Z score is Z = (X - μ)/σ where X represents the raw score, μ represents the mean, and σ represents the standard deviation.

We can rearrange this formula to solve for X:

X = Z*σ + μ

Given that Z = 2.0, μ = 80, and σ = 10, we can plug these values into the formula:

X = 2.0*10 + 80 = 100

Therefore, the raw score that corresponds to a Z score of 2.0 is 100. The answer is choice C.

Question 15

Using a normal curve table,the percentage of scores between a Z score of 1.29 and a Z score of 1.49 is
A)0.54%
B)3.04%
C)7.49%
D)83.34%

Accepted Answer

To find the percentage of scores between a Z score of 1.29 and a Z score of 1.49, we look up the corresponding percentages in a normal distribution table. The percentage of scores less than a Z score of 1.29 is approximately 90.32%, and for a Z score of 1.49, it is approximately 93.32%. Subtracting these two gives us the percentage of scores between these two Z scores: 93.32% - 90.32% = 3.00%. The closest option provided is 3.04%, which accounts for slight variations in normal distribution tables.

Question 16

If the mean score on a stress scale is 5,the standard deviation is 2,and the distribution is normal,the percentage of people who would obtain scores between 5 and 9 is
A)34%
B)14% + 34% = 48%
C)50%
D)34% + 34% = 68%

Accepted Answer

In a normal distribution, the area between the mean and one standard deviation above the mean (5 to 7) accounts for 34% of the data. Extending to two standard deviations above the mean (7 to 9) covers an additional 14%. Therefore, the total percentage of people scoring between 5 and 9 is 34% + 14% = 48%.

Question 17

Using a normal curve table,if a person has a music aptitude score of 41,which equals a Z score of 1.3,the percentage of people having a higher score is
A)40.32% − 50% = -9.68%
B)50% − 40.32% = 9.68%
C)100% − 40.32% = 59.68%
D)50% + 40.32% = 90.32%

Accepted Answer

Using a normal curve table, we find that a Z score of 1.3 corresponds to an area of 0.9032. This means that 90.32% of people have a music aptitude score lower than 41. To find the percentage of people with a higher score, we subtract this value from 50% (since the normal curve is symmetrical).

50% - 40.32% = 9.68%

Therefore, the percentage of people with a higher score is 9.68%.

Question 18

Using a normal curve table,if a person has a score of 4.78 on a test,which equals a Z score of 1.5,the percentage of cases that lie above this score is
A)43.32% − 50% = -6.68%
B)50% − 43.32% = 6.68%
C)100% − 43.32% = 56.68%
D)50% + 43.32% = 93.32%

Accepted Answer

Using a normal curve table, we can find that a Z score of 1.5 corresponds to a percentage of 93.32% that lies below that score. Therefore, to find the percentage of cases that lie above the score of 4.78, we subtract 93.32% from 100%, which gives 6.68% (Option B).

Question 19

If the mean score on a stress scale is 5,the standard deviation is 2,and the distribution is normal,what would be the mean value when converted to a Z distribution?
A)0
B)2
C)5
D)not enough information to make a determination

Accepted Answer

In a Z distribution, the mean is always 0 because Z-scores represent the number of standard deviations a score is from the mean.

Question 20

Which of the following statements regarding Z scores is FALSE?
A)It indicates whether the raw score is above or below average.
B)It indicates whether or not a distribution is skewed.
C)It indicates the percentage of cases between that score and the mean.
D)It is sometimes referred to as a standard score.

Accepted Answer

Z scores are primarily used to determine how far and in what direction a score deviates from the mean in standard deviation units, not to indicate whether a distribution is skewed. Skewness is a measure of the asymmetry of the probability distribution of a real-valued random variable about its mean, and while extreme scores might suggest potential skewness, Z scores themselves do not directly indicate skewness.

Quiz 3: Some Key Ingredients for Inferential Statistics: Z Scores,the Normal Curve, Sample Versus Population, and Probability

Using a normal curve table,the Z score a person would need to be in the top 4% of his or her class on a particular test is

Using a normal curve table,if a person has a score in the bottom 42% of all scores in a distribution,the highest Z score this person can have is

Using a normal curve table,if a person received a test score that is in the top 32% of all test scores,the person's Z score must be at least

In a distribution of Z scores,the mean is always

If the mean score on a creativity test is 20,the standard deviation is 5,and the distribution is normal,the percentage of people who would obtain scores between 15 and 25 is

The normal curve is most likely to be described as

A psychologist is most likely to calculate Z scores in order to

If the mean of a test is 13 and its standard deviation is 3,the Z score for a person with a raw score of 7 is

Using the percentage approximations for the normal curve,the percentage of scores between the mean and one standard deviation below the mean is

The variance of a distribution of Z scores is always

The raw score that corresponds to a Z score of 2.0 obtained from a distribution with a mean of 80 and a standard deviation of 10 is

Using a normal curve table,the percentage of scores between a Z score of 1.29 and a Z score of 1.49 is

If the mean score on a stress scale is 5,the standard deviation is 2,and the distribution is normal,the percentage of people who would obtain scores between 5 and 9 is

Using a normal curve table,if a person has a music aptitude score of 41,which equals a Z score of 1.3,the percentage of people having a higher score is

Using a normal curve table,if a person has a score of 4.78 on a test,which equals a Z score of 1.5,the percentage of cases that lie above this score is

If the mean score on a stress scale is 5,the standard deviation is 2,and the distribution is normal,what would be the mean value when converted to a Z distribution?

Which of the following statements regarding Z scores is FALSE?