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Difficulty: HardNormal Distributions, Standard Deviation Curves, and Percentile Ranks

The distribution of scores on Exam X is normally distributed with a mean of 7070 and a standard deviation of 88. The distribution of scores on Exam Y is normally distributed with a mean of 150150 and a standard deviation of 2020. Candidate Score AA is on Exam X and has a z-score of +1.5+1.5. Candidate Score BB is on Exam Y and corresponds to the 84th percentile of Exam Y.

Which of the following statements must be true? Select all such statements.

  1. Candidate Score AA is equal to 8282.Answer
  2. The proportion of test-takers on Exam Y scoring higher than Candidate Score BB is approximately 16%16\%.Answer
  3. C
    A score of 6262 on Exam X corresponds to a higher percentile rank than Candidate Score BB on Exam Y.
  4. D
    A score of 8686 on Exam X corresponds to the 99.7th percentile of Exam X.
  5. A score of 110110 on Exam Y has a z-score of −2.0-2.0.Answer

Answer

The statements asserting that Candidate Score AA is equal to 8282, that approximately 16%16\% of test-takers score higher than Candidate Score BB, and that a score of 110110 on Exam Y has a z-score of −2.0-2.0 are all correct.
The correct statements correctly apply the z-score formula and standard normal curve benchmarks: Candidate Score AA is 8282 because 70+1.5(8)=8270 + 1.5(8) = 82; the proportion of scores above the 84th percentile is 100%−84%=16%100\% - 84\% = 16\%; and a score of 110110 on Exam Y is 4040 points below the mean of 150150, giving a z-score of −4020=−2.0\frac{-40}{20} = -2.0.

Step-by-Step Solution

1
Calculate Candidate Score AA on Exam X using the z-score formula.
A=μ+z⋅σ=70+(1.5)(8)=82A = \mu + z \cdot \sigma = 70 + (1.5)(8) = 82.
A z-score of +1.5+1.5 means the score is 1.51.5 standard deviations above the mean.
2
Analyze Candidate Score BB at the 84th percentile of Exam Y.
Percentile rank of 84th means 84%84\% score below or equal to BB, so 100%−84%=16%100\% - 84\% = 16\% score higher than BB. Under the standard normal curve (empirical rule), the 84th percentile corresponds to z=+1.0z = +1.0, which gives B=150+1.0(20)=170B = 150 + 1.0(20) = 170.
Percentile rank measures the proportion of values below a given point.
3
Evaluate the percentile rank of a score of 6262 on Exam X.
z=62−708=−1.0z = \frac{62 - 70}{8} = -1.0. A z-score of −1.0-1.0 corresponds to the 16th percentile (50%−34%=16%50\% - 34\% = 16\%).
Since the 16th percentile is less than the 84th percentile (Candidate Score BB), this statement is false.
4
Evaluate the percentile rank of a score of 8686 on Exam X.
z=86−708=+2.0z = \frac{86 - 70}{8} = +2.0. A z-score of +2.0+2.0 corresponds to the 97.5th percentile (50%+34%+13.5%=97.5%50\% + 34\% + 13.5\% = 97.5\%).
The 99.7th percentile corresponds to z=+3.0z = +3.0 (score of 9494), making this statement false.
5
Calculate the z-score for a score of 110110 on Exam Y.
z=110−15020=−2.0z = \frac{110 - 150}{20} = -2.0.
The raw score 110110 is 4040 units below the mean of 150150, which is exactly 22 standard deviations below the mean.

Key Concept

Normal Distribution, Z-Scores, and Percentile Benchmarks
Estimated Time:2m 0s
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Topics

Normal Distributions, Standard Deviation Curves, and Percentile Ranks

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