Chapter 5 Quiz Part 2.
A distribution of scores for a test of life stressors has a mean of µ = 125 and standard deviation of σ = 15. The researcher calculates z-score to standardize the distribution. What are the mean and the standard deviation for the z-score in this distribution? a. µ = 100, σ = 25 b. µ = 0, σ = 10 c. µ = 0, σ = 1 d. µ = 50, σ = 1
C
The distribution for scores on a history exam has µ = 70 and σ = 4. The distribution for a set of scores on a sociology exam has µ = 68 and σ = 4. Tenisha scored 76 on both exams. Which of the following statements best describes the grades Tenisha will receive on her exams? a. She will receive the same grades on both exams because her raw scores are the same. b. She will receive the same grades on both exams because σ is the same for both distributions. c. She will receive a higher grade on the history exam based on standardization. d. She will receive a higher grade on the sociology exam based on standardization.
D
A distribution of exam scores has a mean of µ = 52 and standard deviation of σ = 6. What is the z-score for the exam score X = 61? a. z = +1.5 b. z = −1.5 c. z = +.50 d. z = +2.0
A
What is necessary to determine the z-score for a raw score in a particular set of scores? a. The highest, lowest, and mean scores for the set b. The number of scores and standard deviation for the set c. The median X score and the mean for the set d. The mean and standard deviation for the set
D
For a population with µ = 69 and σ = 4, which of the following X scores will convert to a positive z-score of magnitude greater than 1.0? a. X = 72 b. X = 65 c. X = 74 d. X = 62
C
When Lorenzo finished his exam last week, he thought the test was over. But the instructor put z-score on each student's paper and asked them to figure out their original score. The mean for the class is µ = 61 and standard deviation is σ = 8. Lorenzo's z-score is +1.75. What did he score on the exam? a. 53 b. 69 c. 75 d. 47
C
Within a population having standard deviation of σ = 20, the raw score X = 75 has a z-score of −1.25. What is the mean for this population? a. µ = 90 b. µ = 125 c. µ = 100 d. µ = 50
C
Professor Chao's engineering final exam has a mean of µ = 79 and σ = 7. After standardizing the exam scores, the mean is µ = 80 and σ = 5. How did Professor Chao arrive at these new standardized scores for the exam? a. Deciding what values will be simple to calculate b. Comparing z-scores across all exams in the college c. Selecting the nearest values that are multiples X d. Factoring in how many students took the exam
A
A distribution with µ = 61 and σ = 6 has been standardized to reflect a new mean of µ = 50 and a standard deviation of σ = 10. What is the new standardized score for a score of X = 52 in the original distribution? a. X = 65 b. X = 70 c. X = 40 d. X = 35
D
As part of a clinical research project, a sample population has been using an experimental drug intended to improve memory. Study participants completed a comprehensive memory test following six weeks of treatment. The results were standardized and compared to mean results for the full population. Which of the following participants indicates promising results for this drug treatment? a. Liming, a 34-year-old woman with a memory test z-score of +1.25 b. Ahmad, a 61-year-old man with a memory test z-score of +.50 c. Javier, a 48-year-old man with a memory test z-score of −.25 d. Ellie, a 52-year-old woman with a memory test z-score of +2.75
D
A researcher has compiled a set of scores with a mean of µ = 39 and standard deviation of σ = 4. What is the z-score for the raw score X = 28? a. z = −2.75 b. z = +.60 c. z = −.50 d. z = +2.75
A
A sample has a mean of M= 64 and standard deviation of s= 3. What is the z-score for a sample score of X = 70? a. z = −.75 b. z = +2.00 c. z = +1.25 d. z = +1.00
B
A distribution has µ = 50 and σ = 6. In a graphical representation of the distribution, where would the score X = 20 appear? a. In the left tail of the curve b. Just left of the peak in the curve c. Just right of the peak in the curve d. In the right tail of the curve
A
A calculus exam has a mean of µ = 73 and a standard deviation of σ = 4. Trina's score on the exam was 79, giving her a z-score of +1.50. The teacher standardized the exam distribution to a new mean of µ = 70 and standard deviation of σ = 5. What is Trina's z-score for the standardized distribution of the calculus exam? a. z = +.50 b. z = +2.25 c. z = +.75 d. z = +1.50
D
In a distribution, a score of X = 60 has a z-score of −3.0. A score of X = 85 has a z-score of +2.0. What are the mean and the standard deviation for this population? a. µ = 70, σ = 2 b. µ = 50, σ = 10 c. µ = 68, σ = 4 d. µ = 75, σ = 5
D
A distribution of 1000 test scores has a mean of µ = 287. Following standardization, what is the standard deviation for this distribution? a. σ = 25 b. σ =14 c. σ = 1 d. Not enough data to calculate
C
A test of life satisfaction has a mean of µ = 60. Jenna's test score of X = 72 has a z-score of +3.0. What is the standard deviation for this distribution of test scores? a. σ = 4 b. σ = 3 c. σ = 6 d. σ = 2
A
What is the benefit of converting raw scores from a sample into z-scores? a. The new scores are easier to compare. b. The new scores are all relative to 100. c. The new scores are all higher than zero. d. The new scores are easier to average.
A
Which of the following z-scores indicates an X value that falls farthest below the mean for the distribution? a. z = −1.50 b. z = −.50 X c. z = +2.00 d. z = +1.25
A
Marisol received a score of X = 70 on her chemistry exam and a score of X = 59 on her geography exam. The mean for the chemistry exam is µ = 66, and the mean for the geography exam is µ = 50. On which exam will Marisol receive a higher grade? a. Unknown; the total number of raw scores for each exam is required. b. Geography; this score is higher above the mean than her chemistry score. c. Chemistry; her exam score of 70 is higher than the geography score of 59. d. Unknown; the standard deviation for each set of exam scores is required.
D
