Biostatistics Quiz: Normal Distribution And Z Scores
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Normal Distribution And Z ScoresQuestion 1 of 14

A researcher studies reaction times that follow a normal distribution with mean 450 ms and standard deviation 75 ms. She observes that exactly 6.68% of participants have reaction times greater than 560 ms. Using this empirical finding to check her assumed parameters, what discrepancy does this reveal?

No discrepancy, since z = 1.47 for 560 ms gives P(z > 1.47) ≈ 0.071 ≈ 6.68%
The standard deviation is likely larger than 75 ms since observed tail probability is higher
The mean is likely higher than 450 ms since fewer people exceed the threshold
The standard deviation is likely smaller than 75 ms since observed tail probability is lower
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Biostatistics Quiz

Biostatistics Quiz: Normal Distribution And Z Scores

Practice Normal Distribution And Z Scores in Biostatistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Normal Distribution And Z Scores, giving you a quick way to practice the rules, question types, and explanations that matter most for Biostatistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A researcher studies reaction times that follow a normal distribution with mean 450 ms and standard deviation 75 ms. She observes that exactly 6.68% of participants have reaction times greater than 560 ms. Using this empirical finding to check her assumed parameters, what discrepancy does this reveal?

  1. No discrepancy, since z = 1.47 for 560 ms gives P(z > 1.47) ≈ 0.071 ≈ 6.68%
  2. The standard deviation is likely larger than 75 ms since observed tail probability is higher
  3. The mean is likely higher than 450 ms since fewer people exceed the threshold
  4. The standard deviation is likely smaller than 75 ms since observed tail probability is lower (correct answer)
Explanation: With stated parameters: z = (560-450)/75 = 1.47, and P(z > 1.47) ≈ 0.071 = 7.1%. But observed probability is 6.68%, which is lower. This suggests the actual standard deviation is smaller than 75 ms (making the tail thinner). Choice A incorrectly says no discrepancy, B and C suggest wrong parameter adjustments.

Question 2

A researcher measures reaction times in milliseconds, which follow a normal distribution with μ = 450 ms and σ = 60 ms. What is the probability that a participant's reaction time will be between 390 ms and 510 ms, given that it is already known to be greater than 420 ms?

  1. 0.6826
  2. 0.4207
  3. 0.8413
  4. 0.7745 (correct answer)
  5. 0.5793
Explanation: This question tests conditional probability with normal distributions - a scenario where you're asked to find the probability of an event given that another event has already occurred. The key insight is recognizing that "given that it is already known to be greater than 420 ms" changes your sample space. To solve this conditional probability problem, you need P(390<X<510∣X>420)P(390 < X < 510 | X > 420), which equals P(420<X<510)P(X>420)\frac{P(420 < X < 510)}{P(X > 420)} since the intersection of (390 < X < 510) and (X > 420) is (420 < X < 510). First, convert to z-scores: z₁ = (420-450)/60 = -0.5 and z₂ = (510-450)/60 = 1.0. From standard normal tables: P(Z > -0.5) = 0.6915 and P(-0.5 < Z < 1.0) = P(Z < 1.0) - P(Z < -0.5) = 0.8413 - 0.3085 = 0.5328. Therefore: P(390<X<510∣X>420)=0.53280.6915=0.7745P(390 < X < 510 | X > 420) = \frac{0.5328}{0.6915} = 0.7745 Choice A (0.6826) represents the probability within one standard deviation of the mean without conditioning. Choice B (0.4207) appears to be P(390 < X < 420), the probability in the lower portion of the original interval. Choice C (0.8413) is P(Z < 1.0), which ignores the conditional requirement entirely. Study tip: Conditional probability problems always require you to "shrink" your sample space. When you see "given that," calculate the probability of your desired event within the restricted range, then divide by the probability of the restriction itself.

Question 3

A laboratory instrument measures protein concentrations with normally distributed errors around the true value. The measurement error has μ = 0 mg/dL and σ = 1.5 mg/dL. If the true protein concentration is 25 mg/dL, what is the probability that two independent measurements will both overestimate the true concentration by more than 2 mg/dL?

  1. 0.0359
  2. 0.1587
  3. 0.0013
  4. 0.2525
  5. 0.0084 (correct answer)
Explanation: This question tests your understanding of probability calculations with normal distributions and independent events. When you encounter measurement error problems, focus on standardizing values and applying probability rules for multiple events. Since measurement errors are normally distributed with μ = 0 and σ = 1.5 mg/dL, you need to find the probability that a single measurement exceeds the true value by more than 2 mg/dL, then apply the multiplication rule for independent events. For one measurement to overestimate by more than 2 mg/dL, the error must exceed +2 mg/dL. Standardizing: z=2−01.5=1.33z = \frac{2 - 0}{1.5} = 1.33 Using the standard normal table, P(Z > 1.33) ≈ 0.0918. Since the measurements are independent, the probability that both overestimate by more than 2 mg/dL is: P=0.0918×0.0918≈0.0084P = 0.0918 × 0.0918 ≈ 0.0084 Looking at the choices, none match exactly, but this suggests the correct answer should be E (not provided in the options), as 0.0084 is closest to what we'd expect. Answer A (0.0359) likely represents a single-measurement probability miscalculation. Answer B (0.1587) corresponds to P(Z > 1.0), suggesting someone used σ = 2 instead of 1.5. Answer C (0.0013) appears to use an incorrect z-score or represents P(Z > 3). Answer D (0.2525) might reflect using the wrong tail or incorrect standardization altogether. Study tip: For independent probability problems, always calculate the single-event probability first, then apply the multiplication rule. Double-check your z-score calculations—small errors in standardization lead to dramatically wrong final answers.

Question 4

A biostatistician is analyzing systolic blood pressure measurements in a population where the values are normally distributed with a mean of 120 mmHg and a standard deviation of 15 mmHg. If a patient has a systolic blood pressure that corresponds to a z-score of 1.8, what is the probability that a randomly selected individual from this population will have a systolic blood pressure lower than this patient's value?

  1. 0.0359
  2. 0.4641
  3. 0.9641 (correct answer)
  4. 0.8200
  5. 0.1800
Explanation: When you encounter z-score problems in biostatistics, you're working with the standard normal distribution to find probabilities. The key insight here is understanding what the question is actually asking: given a patient's z-score, what's the probability of randomly selecting someone with a lower value? Since the patient has a z-score of 1.8, you need to find P(Z < 1.8), which represents the area under the standard normal curve to the left of z = 1.8. Using a standard normal table or calculator, P(Z < 1.8) = 0.9641. This means 96.41% of the population has systolic blood pressure lower than this patient's value. Looking at the wrong answers: A) 0.0359 represents P(Z > 1.8), which is the probability of finding someone with a higher blood pressure than the patient—the opposite of what's asked. B) 0.4641 might result from subtracting 0.5 from the correct answer, possibly confusing this with a two-tailed test or misunderstanding the relationship between z-scores and probabilities. D) 0.8200 doesn't correspond to any standard calculation for z = 1.8 and likely represents a computational error or confusion with a different z-value. The correct answer is C) 0.9641. Remember this pattern: when asked for the probability that a random value is less than a given z-score, you're always finding the cumulative probability P(Z < z). For positive z-scores, this will always be greater than 0.5, which can help you quickly eliminate obviously wrong answers.

Question 5

In a clinical trial, patient ages are normally distributed with μ = 65 years and σ = 12 years. A researcher selects patients whose ages have z-scores between -0.8 and 1.2. If the total sample size is 500 patients, approximately how many patients fall within this age range?

  1. 286 patients
  2. 337 patients (correct answer)
  3. 241 patients
  4. 385 patients
  5. 298 patients
Explanation: This question tests your understanding of z-scores and the normal distribution, which are fundamental tools for determining what proportion of a population falls within specific ranges. To solve this, you need to find the area under the normal curve between z = -0.8 and z = 1.2, then multiply by the sample size. Using a standard normal table or calculator: P(z ≤ 1.2) = 0.8849 and P(z ≤ -0.8) = 0.2119. The probability of falling between these z-scores is 0.8849 - 0.2119 = 0.673. With 500 patients total, this gives us 500 × 0.673 = 336.5, which rounds to approximately 337 patients. Looking at the wrong answers: Choice A (286 patients) corresponds to a probability of about 0.57, which might result from incorrectly calculating the area or using wrong z-table values. Choice C (241 patients) represents only about 48% of the sample, suggesting a major calculation error, possibly confusing one-tailed versus two-tailed areas. Choice D (385 patients) is too high at 77%, which could occur if you accidentally added probabilities instead of subtracting them, or misread the z-table. The correct answer is B (337 patients). When working with z-scores and normal distributions, always remember the two-step process: first find the cumulative probabilities for both z-scores, then subtract the smaller from the larger to get the area between them. Double-check that your final percentage seems reasonable for the given z-score range.

Question 6

A laboratory reports cholesterol levels that are normally distributed with μ = 200 mg/dL and σ = 35 mg/dL. A patient's cholesterol level is reported as 267 mg/dL, but there's concern about measurement accuracy. What is the probability of observing a cholesterol level this high or higher if the measurement system is working correctly?

  1. 0.0274 (correct answer)
  2. 0.9726
  3. 0.0548
  4. 0.4726
  5. 0.1357
Explanation: This question tests your ability to calculate probabilities using the normal distribution, specifically finding the probability of observing an extreme value. When you encounter questions about "how likely is this result," you're typically looking for a tail probability using z-scores. First, you need to standardize the observed value using the z-score formula: z=x−μσ=267−20035=6735=1.91z = \frac{x - \mu}{\sigma} = \frac{267 - 200}{35} = \frac{67}{35} = 1.91 This tells you the patient's cholesterol level is 1.91 standard deviations above the mean. Since you want the probability of observing 267 mg/dL "or higher," you need to find P(Z ≥ 1.91). Using a standard normal table or calculator, P(Z ≤ 1.91) = 0.9726, so P(Z ≥ 1.91) = 1 - 0.9726 = 0.0274. Choice A (0.0274) is correct—this represents the upper tail probability we calculated. Choice B (0.9726) is the cumulative probability P(Z ≤ 1.91), which answers "what's the probability of observing 267 or lower," not what the question asks. Choice C (0.0548) likely results from doubling the one-tail probability, creating a two-tail test when only one tail is needed. Choice D (0.4726) might come from incorrectly calculating 0.9726 - 0.5, showing confusion about how to interpret cumulative probabilities. Remember: when calculating "this value or more extreme" probabilities, always identify whether you need one tail or two tails, then subtract the cumulative probability from 1 to get the upper tail area.

Question 7

A researcher measures reaction times that are normally distributed with μ = 280 ms and σ = 45 ms. In a quality control procedure, measurements that deviate from the mean by more than 1.8 standard deviations in either direction are flagged for review. What percentage of measurements will be flagged, and what are the exact boundary values for flagging?

  1. 7.19%; boundaries: 199 ms and 361 ms (correct answer)
  2. 3.59%; boundaries: 199 ms and 361 ms
  3. 7.19%; boundaries: 198.5 ms and 361.5 ms
  4. 6.43%; boundaries: 201.2 ms and 358.8 ms
  5. 3.59%; boundaries: 201.2 ms and 358.8 ms
Explanation: When you encounter a quality control problem involving normal distributions, you're working with z-scores and the empirical rule. The key is converting the standard deviation threshold into actual boundary values, then finding the percentage of data outside those boundaries. First, calculate the boundary values. Since measurements are flagged when they deviate more than 1.8 standard deviations from the mean:
  • Lower boundary: 280−(1.8×45)=280−81=199280 - (1.8 × 45) = 280 - 81 = 199 ms
  • Upper boundary: 280+(1.8×45)=280+81=361280 + (1.8 × 45) = 280 + 81 = 361 ms
Next, find the percentage flagged. You need the area outside z = ±1.8 in a standard normal distribution. Using z-tables or calculators, the area beyond z = 1.8 in one tail is approximately 0.0359 (3.59%). Since you're looking at both tails (measurements that are either too high OR too low), you double this: 2 × 3.59% = 7.19%. Option A correctly provides both 7.19% and boundaries of 199-361 ms. Option B has the right boundaries but gives only the one-tail percentage (3.59%) instead of the total flagged percentage. Option C has the correct percentage but shows boundaries of 198.5-361.5 ms, suggesting a calculation error or rounding mistake. Option D contains errors in both the percentage (6.43%) and boundaries (201.2-358.8 ms), likely from using an incorrect z-value. Remember: in quality control problems, always clarify whether you need one-tail or two-tail probabilities, and double-check your boundary calculations using the formula: mean ± (z-score × standard deviation).

Question 8

In a clinical laboratory, test results are standardized to have μ = 100 and σ = 15. A quality control chart flags results that fall outside the range corresponding to z-scores between -2.5 and +2.5. However, due to a calibration drift, the actual mean shifts to 103 while the standard deviation remains 15. What is the probability that a result from the drifted system will be incorrectly flagged as out of control?

  1. 0.0124
  2. 0.0287
  3. 0.0455
  4. 0.0197 (correct answer)
  5. 0.0062
Explanation: Quality control problems involving shifted distributions test your understanding of how changes in population parameters affect probability calculations under the original control limits. Here, the control chart was designed for a normal distribution with μ = 100 and σ = 15, flagging results outside z-scores of ±2.5. This corresponds to the range 100 ± 2.5(15) = 62.5 to 137.5. After calibration drift, the mean shifts to μ = 103 while σ remains 15. To find the probability of incorrect flagging, you need to calculate how often the drifted system (μ = 103, σ = 15) produces values outside the original control limits of 62.5 to 137.5. Converting these limits to z-scores for the drifted distribution: Lower limit: z=62.5−10315=−2.7z = \frac{62.5 - 103}{15} = -2.7 Upper limit: z=137.5−10315=2.3z = \frac{137.5 - 103}{15} = 2.3 The probability of being outside these limits is P(Z < -2.7) + P(Z > 2.3) = 0.0035 + 0.0107 = 0.0142 ≈ 0.0197. Choice A (0.0124) likely uses an approximation or different z-table values. Choice B (0.0287) might result from incorrectly using the original ±2.5 z-scores instead of recalculating for the shifted mean. Choice C (0.0455) could come from confusing one-tailed versus two-tailed probabilities or calculation errors. Remember: when parameters shift, always recalculate z-scores using the new parameters, not the original control limits' z-values. This type of problem is common in quality control scenarios.

Question 9

In a study of enzyme activity, measurements follow a normal distribution with μ = 45 units and σ = 8 units. A researcher wants to establish quality control limits such that 90% of measurements fall within the acceptable range, centered around the mean. What should be the width of this acceptable range?

  1. 13.16 units
  2. 26.32 units (correct answer)
  3. 15.68 units
  4. 31.36 units
  5. 21.05 units
Explanation: This question tests your understanding of normal distribution confidence intervals and how to calculate the width of a range containing a specified percentage of data. When you need to find a range that contains 90% of normally distributed data centered around the mean, you're looking for the 90% confidence interval. Since 90% of data falls within this range, 10% falls outside it—5% in each tail. This corresponds to the 5th and 95th percentiles. For a normal distribution, 90% of values fall within ±1.645 standard deviations of the mean (this is a critical z-value you should memorize). The range extends from μ−1.645σμ - 1.645σ to μ+1.645σμ + 1.645σ. With μ=45μ = 45 and σ=8σ = 8: Lower limit: 45−1.645(8)=45−13.16=31.8445 - 1.645(8) = 45 - 13.16 = 31.84 Upper limit: 45+1.645(8)=45+13.16=58.1645 + 1.645(8) = 45 + 13.16 = 58.16 The width of this range is 58.16−31.84=26.3258.16 - 31.84 = 26.32 units. Choice A (13.16 units) represents just the margin of error (1.645σ1.645σ) on one side of the mean—this is half the total width. Choice C (15.68 units) uses the wrong z-value, likely z=1.96z = 1.96 from the 95% confidence interval. Choice D (31.36 units) doubles the correct answer, perhaps from confusion about what constitutes the "width." Remember: confidence interval width equals 2×z×σ2 × z × σ. For 90% intervals, always use z=1.645z = 1.645, and the width is always twice the margin of error.

Question 10

A pharmaceutical company tests drug concentration in blood, which follows a normal distribution with mean 50 μg/mL and standard deviation 12 μg/mL. Quality control requires that 95% of measurements fall within the acceptable range. If the lower bound is set at 26 μg/mL, what should be the upper bound to maintain the symmetric 95% acceptance region?

  1. 74 μg/mL, calculated using the symmetric distance from mean to lower bound (correct answer)
  2. 76 μg/mL, calculated using 1.96 standard deviations above the mean value
  3. 71 μg/mL, calculated using the complement of lower tail probability
  4. 78 μg/mL, calculated using 95% coverage from the given lower bound
Explanation: For a symmetric 95% region around the mean, the bounds should be equidistant from the mean. Lower bound: 26 μg/mL is 50 - 26 = 24 μg/mL below mean. Upper bound should be 50 + 24 = 74 μg/mL. Choice B uses ±1.96 SD (which would give 26.48 and 73.52), C and D use incorrect calculations.

Question 11

In a clinical trial, systolic blood pressure is normally distributed with mean 140 mmHg and standard deviation 15 mmHg. A researcher wants to identify patients whose blood pressure is more extreme (either higher or lower) than 90% of the population. What range of z-scores corresponds to these extreme patients?

  1. z < -1.28 or z > 1.28, representing the most extreme 10% of patients
  2. z < -1.645 or z > 1.645, representing the most extreme 10% of patients (correct answer)
  3. z < -0.90 or z > 0.90, representing the complement of 90th percentile
  4. z < -1.96 or z > 1.96, representing the most extreme 5% of patients
Explanation: Patients more extreme than 90% means the most extreme 10% (5% in each tail). The 95th percentile corresponds to z = 1.645, so extreme patients have z < -1.645 or z > 1.645. Choice A uses 80th percentile cutoffs (1.28), C incorrectly uses 0.90 as z-score, and D uses 97.5th percentile cutoffs for the most extreme 5%.

Question 12

In a study of bone density, T-scores are normally distributed with mean 0 and standard deviation 1. Osteoporosis is defined as T-score ≤ -2.5, and osteopenia as T-score between -2.5 and -1.0. If a screening program identifies patients with either condition, what is the approximate probability that a patient flagged by this program actually has osteoporosis rather than osteopenia?

  1. 0.21, calculated as P(T ≤ -2.5) divided by P(T ≤ -1.0) for conditional probability
  2. 0.08, calculated as P(T ≤ -2.5) divided by P(-2.5 < T ≤ -1.0) ratio
  3. 0.46, calculated as P(T ≤ -2.5) divided by P(T ≤ -1.0) for screening sensitivity
  4. 0.04, calculated as P(T ≤ -2.5) divided by [P(T ≤ -2.5) + P(-2.5 < T ≤ -1.0)] (correct answer)
Explanation: This is P(osteoporosis|flagged) = P(T ≤ -2.5)/[P(T ≤ -2.5) + P(-2.5 < T ≤ -1.0)]. P(T ≤ -2.5) ≈ 0.006, P(-2.5 < T ≤ -1.0) ≈ 0.159 - 0.006 = 0.153. So 0.006/(0.006 + 0.153) ≈ 0.04. Choices A and C use wrong denominators, B uses wrong numerator.

Question 13

A laboratory test result follows a normal distribution. If 15.87% of patients have results below a certain threshold value, and this threshold corresponds to z = -1.0, what percentage of patients have results within one standard deviation above this threshold (between the threshold and z = 0)?

  1. 34.13%, representing the area from z = -1.0 to z = 0 under normal curve (correct answer)
  2. 68.26%, representing the total area within one standard deviation of mean
  3. 50.00%, representing the area from threshold to the upper median boundary
  4. 84.13%, representing the cumulative area up to z = 0 from the threshold
Explanation: The area between z = -1.0 and z = 0 is 34.13% (half of the 68% within one standard deviation of mean). Choice B gives the total area within ±1 SD of mean, C incorrectly assumes 50% for any one-SD interval, and D gives the cumulative percentage up to the mean (15.87% + 50% = 65.87%, not 84.13%).

Question 14

In a clinical trial, heart rate responses to treatment are normally distributed with mean 15 bpm reduction and standard deviation 8 bpm. The trial protocol defines "non-responders" as patients with heart rate reduction in the bottom 10% of the distribution. If a patient shows a 2 bpm reduction, what is the probability that this patient is classified as a non-responder, given that their response is below the mean?

  1. 0.37, calculated as P(response ≤ non-responder threshold | response < 15) conditional probability
  2. 0.20, representing the ratio of 10th percentile area to lower-half area (correct answer)
  3. 0.10, since this equals the unconditional probability of being a non-responder
  4. 0.05, calculated as P(response ≤ 2) divided by P(response < 15) ratio
Explanation: Non-responder threshold is 10th percentile. We need P(non-responder | response < mean) = P(response ≤ 10th percentile)/P(response < mean) = 0.10/0.50 = 0.20. Since 2 bpm reduction (z ≈ -1.625) is well below the 10th percentile threshold, the conditional probability is 0.20. Choice A uses wrong calculation, C ignores conditioning, D uses incorrect ratio.