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

The processing times for a specific type of database query on a server cluster are normally distributed with a mean of 180180 milliseconds and a standard deviation of 2525 milliseconds. Which of the following statements must be true? Indicate all such statements.

  1. Approximately 68%68\% of the query processing times are between 155155 milliseconds and 205205 milliseconds.Answer
  2. A query processing time of 230230 milliseconds corresponds to a zz-score of +2.0+2.0.Answer
  3. A query processing time of 205205 milliseconds corresponds to approximately the 84th84\text{th} percentile rank.Answer
  4. D
    A query processing time of 155155 milliseconds corresponds to approximately the 32nd32\text{nd} percentile rank.
  5. E
    A zz-score of +1.0+1.0 indicates that the query processing time is 25%25\% greater than the mean processing time.

Answer

The correct statements are those asserting that approximately 68% of processing times lie between 155 ms and 205 ms, that a processing time of 230 ms corresponds to a z-score of +2.0, and that 205 ms corresponds to approximately the 84th percentile rank.
For a normal distribution with mean μ=180\mu = 180 and standard deviation σ=25\sigma = 25: (1) The interval μ±1σ=[155,205]\mu \pm 1\sigma = [155, 205] contains approximately 68%68\% of all observations. (2) A score of 230230 ms has a zz-score of 230−18025=+2.0\frac{230-180}{25} = +2.0. (3) A score of 205205 ms corresponds to z=+1.0z = +1.0, which accumulates 50%50\% (area below mean) +34%+ 34\% (area between mean and +1σ+1\sigma) = 84%84\% percentile rank.

Step-by-Step Solution

1
Evaluate the 1 standard deviation interval
μ±1σ=180±25=[155,205]\mu \pm 1\sigma = 180 \pm 25 = [155, 205] ms
The 68-95-99.7 empirical rule states that approximately 68% of observations in a normal distribution fall within 1 standard deviation of the mean.
2
Calculate the z-score for a query processing time of 230 ms
z=230−18025=5025=+2.0z = \frac{230 - 180}{25} = \frac{50}{25} = +2.0
The standard score formula z=x−μσz = \frac{x - \mu}{\sigma} measures how many standard deviations an observation is above or below the mean.
3
Determine the percentile rank for 205 ms and 155 ms
For 205 ms (z=+1z = +1), percentile = 50%+34%=84%50\% + 34\% = 84\%. For 155 ms (z=−1z = -1), percentile = 50%−34%=16%50\% - 34\% = 16\%.
50% of the distribution lies below the mean. For z=+1z = +1, add the 34% between mean and z=+1z = +1. For z=−1z = -1, subtract the 34% from 50%.
4
Evaluate relative percentage change versus standard deviation
A 1 standard deviation increase (2525 ms) above the mean (180180 ms) equals an increase of 25180≈13.89%\frac{25}{180} \approx 13.89\%.
Standard deviation measures absolute dispersion in the units of the variable, not a percentage of the mean.

Key Concept

Empirical Rule, z-Score Calculation, and Percentile Ranks in Normal Distributions
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Topics

Normal Distributions, Standard Deviation Curves, and Percentile Ranks

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