Using z-Scores - Math 3

1. A researcher is comparing test performance across two different exams. Exam 1 has a mean of 78 and standard deviation of 12. Exam 2 has a mean of 85 and standard deviation of 15. If Maria scored 90 on Exam 1 and 94 on Exam 2, on which exam did she perform better relative to other test-takers?
   Answer: Exam 1, because her z-score was approximately 1.0 compared to 0.6 on Exam 2 When comparing performance across different exams with different means and standard deviations, you need to use z-scores to standardize the results. Raw scores alone can't tell you relative performance because the exams have different difficulty levels and score distributions. A z-score tells you how many standard deviations above or below the mean a score falls, using the formula: $z = \frac{x - \mu}{\sigma}$ For Maria's Exam 1 performance: $z = \frac{90 - 78}{12} = \frac{12}{12} = 1.0$ For Maria's Exam 2 performance: $z = \frac{94 - 85}{15} = \frac{9}{15} = 0.6$ Maria performed better relative to other test-takers on Exam 1, where she scored 1.0 standard deviation above the mean compared to only 0.6 standard deviations above the mean on Exam 2. Choice A is wrong because the z-scores aren't the same (1.0 vs 0.6). Choice B incorrectly assigns the z-scores to the wrong exams, swapping the values. Choice C falls into the trap of comparing raw scores directly—while 94 is higher than 90, this ignores that Exam 2 had a higher mean and more variability, making Maria's performance less impressive relative to her peers. Remember: when comparing performance across different tests or populations, always calculate z-scores first. Raw scores can be misleading when the underlying distributions differ. The higher z-score indicates better relative performance, regardless of which raw score is larger.
2. In a large manufacturing company, employee salaries are normally distributed with a mean of $65,000 and a standard deviation of $12,000. If Sarah earns $77,000 and Miguel earns $59,000, which statement best describes their relative positions in the salary distribution?
   Answer: Sarah's salary is exactly 1 standard deviation above the mean, while Miguel's salary is exactly 0.5 standard deviations below the mean To find z-scores: For Sarah: z = (77,000 - 65,000) / 12,000 = 12,000 / 12,000 = 1.0. For Miguel: z = (59,000 - 65,000) / 12,000 = -6,000 / 12,000 = -0.5. Choice B incorrectly calculates Miguel's z-score as -1. Choice C uses incorrect arithmetic. Choice D makes calculation errors for both employees.
3. Two athletes compete in different sports with normally distributed performance metrics. Athlete X has a z-score of 1.8 in swimming, while Athlete Y has a z-score of -2.1 in track. A sports analyst wants to determine which athlete's performance is more extreme relative to their respective sport's typical performance. What should the analyst conclude?
   Answer: Athlete Y is more extreme because |-2.1| = 2.1 > |1.8| = 1.8, indicating greater distance from the mean To determine which performance is more extreme, we compare the absolute values of the z-scores. |1.8| = 1.8 for Athlete X and |-2.1| = 2.1 for Athlete Y. Since 2.1 > 1.8, Athlete Y's performance is more extreme (further from the mean). Choice A incorrectly focuses on the sign rather than magnitude. Choice C introduces irrelevant information about sport difficulty. Choice D incorrectly claims they're equally extreme.
4. In a psychology experiment, reaction times are normally distributed with mean 800 ms and standard deviation 120 ms. A researcher observes that Participant 1's z-score is exactly twice that of Participant 2's z-score. If Participant 2 had a reaction time of 920 ms, what was Participant 1's reaction time?
   Answer: 1040 ms, because doubling Participant 2's z-score of 1.0 gives z = 2.0 for Participant 1 First find Participant 2's z-score: z₂ = (920-800)/120 = 120/120 = 1.0. Participant 1's z-score is twice this: z₁ = 2(1.0) = 2.0. Convert back to reaction time: time = 800 + 2.0(120) = 800 + 240 = 1040 ms. Choice B has an incorrect time calculation. Choice C incorrectly assumes the doubled z-score should be negative. Choice D makes an unjustified assumption about opposite direction.
5. A standardized test is administered to two different populations. Population A has mean score 480 with standard deviation 60, while Population B has mean score 520 with standard deviation 40. A student from Population A scored 540, and a student from Population B scored 580. Which conclusion about their relative standings is most accurate?
   Answer: The Population B student performed better relative to their group, with a z-score of 1.5 versus 1.0 for the Population A student Population A student: z = (540-480)/60 = 60/60 = 1.0. Population B student: z = (580-520)/40 = 60/40 = 1.5. The higher z-score (1.5) indicates better relative performance. Choice A reverses the z-score assignments. Choice C incorrectly calculates both z-scores as 1.2. Choice D incorrectly compares raw scores instead of relative performance within each population.
6. A coffee shop tracks daily sales, which are normally distributed with mean $2400 and standard deviation $300. On Monday, sales were $2100. On Tuesday, sales increased by $450. How do the z-scores for these two days compare?
   Answer: Monday: z = -1.0, Tuesday: z = 0.5, so Monday's sales were more unusual than Tuesday's Monday's z-score: z = (2100-2400)/300 = -300/300 = -1.0. Tuesday's sales: 2100 + 450 = 2550. Tuesday's z-score: z = (2550-2400)/300 = 150/300 = 0.5. Since |-1.0| = 1.0 > |0.5| = 0.5, Monday was more unusual. Choice A reaches the wrong conclusion despite correct calculations. Choice C incorrectly calculates Tuesday's z-score. Choice D has incorrect Tuesday z-score and wrong conclusion about equal unusualness.
7. Two normally distributed datasets have the same mean but different standard deviations. Dataset X has SD = 10, Dataset Y has SD = 5. If a value in Dataset X has z-score 1.2, what z-score would the same raw value have in Dataset Y?
   Answer: z = 2.4, because the z-score doubles when the standard deviation is halved If the value has z-score 1.2 in Dataset X, it's 1.2(10) = 12 units above the mean. In Dataset Y with the same mean but SD = 5, this same raw value would have z-score = 12/5 = 2.4. Choice A incorrectly reduces the z-score when SD decreases. Choice C incorrectly claims z-scores are independent of SD. Choice D has the right answer but incomplete reasoning about the relationship.
8. Two different normal distributions are being compared. Distribution A has μ = 80 and σ = 15, while Distribution B has μ = 120 and σ = 20. A value of 95 from Distribution A and a value of 140 from Distribution B are being compared using z-scores. Which statement is correct?
   Answer: Both values have identical z-scores and represent equally extreme positions relative to their distributions For Distribution A: z = (95 - 80)/15 = 15/15 = 1.0. For Distribution B: z = (140 - 120)/20 = 20/20 = 1.0. Both z-scores are identical at 1.0, meaning both values are exactly 1 standard deviation above their respective means. Choice A incorrectly assumes Distribution A has a higher z-score. Choice B incorrectly assumes Distribution B has a higher z-score. Choice D correctly identifies equal extremeness but incorrectly states Distribution A has a lower z-score.
9. A pharmaceutical company tests reaction times for a new medication. Reaction times are normally distributed with mean 450 milliseconds and standard deviation 60 milliseconds. If a patient has a reaction time that corresponds to a z-score of -1.5, and another patient's reaction time is 30 milliseconds faster, what is the z-score for the second patient?
   Answer: $z = -2.0$ because the second patient's time is 30 milliseconds faster in a distribution with SD = 60 First, find the first patient's reaction time: if z = -1.5, then time = 450 + (-1.5)(60) = 360 ms. The second patient is 30 ms faster, so their time is 360 - 30 = 330 ms. The z-score is (330-450)/60 = -120/60 = -2.0. Choice B incorrectly calculates the effect of the 30 ms change. Choice C has the right z-score but wrong reasoning about the time calculation. Choice D has incorrect time and z-score values.
10. A student's test scores in three different classes are normally distributed. In Math, she scored 85 (mean = 78, SD = 4). In Science, she scored 92 (mean = 88, SD = 3). In English, she scored 76 (mean = 70, SD = 2.5). If colleges use z-scores to compare performance across subjects, in which subject did she perform best relative to her peers?
    Answer: English, because her z-score of 2.4 shows she performed better relative to peers than in other subjects Calculate z-scores for each subject. Math: z = (85-78)/4 = 1.75. Science: z = (92-88)/3 = 1.33. English: z = (76-70)/2.5 = 2.4. The highest z-score indicates best relative performance, which is English at 2.4. Choice A correctly calculates Math's z-score but incorrectly concludes it's highest. Choice B has the correct Science z-score but wrong conclusion. Choice D has an incorrect Science z-score calculation.
11. A medical study tracks blood pressure readings that are normally distributed with mean 120 mmHg and standard deviation 15 mmHg. Patient X has a reading with z-score 1.6. Patient Y has a reading that is 18 mmHg higher than Patient X. What can be concluded about Patient Y's z-score?
    Answer: Patient Y's z-score is 2.8 When you encounter z-score problems involving multiple patients or data points, remember that z-scores measure how many standard deviations away from the mean a value falls. The key insight is understanding how changes in raw values translate to changes in z-scores. First, let's find Patient X's blood pressure reading. Using the z-score formula: $z = \frac{x - \mu}{\sigma}$, we get $1.6 = \frac{x - 120}{15}$. Solving: $x = 120 + (1.6)(15) = 144$ mmHg. Patient Y's reading is 18 mmHg higher than Patient X's, so Patient Y has $144 + 18 = 162$ mmHg. Now we calculate Patient Y's z-score: $z_Y = \frac{162 - 120}{15} = \frac{42}{15} = 2.8$. Looking at the wrong answers: Choice A (3.6) incorrectly adds the z-scores: $1.6 + 1.8 = 3.4$ (close to 3.6), but z-scores don't add this way. Choice B (2.2) might come from adding $1.6 + 0.6$, perhaps confusing the 18 mmHg increase with a smaller z-score change. Choice C (1.8) represents just the change in z-score ($\frac{18}{15} = 1.2$, though this calculation is also flawed), not Patient Y's actual z-score. The correct answer is D: Patient Y's z-score is 2.8. Study tip: When dealing with z-score changes, always convert back to raw values first, make the specified change, then recalculate the new z-score. Don't try to directly manipulate z-scores unless you're certain about the mathematical relationship.
12. A standardized test has scores that are normally distributed with a mean of 500 and a standard deviation of 100. If a student's z-score is -1.5, and another student scored 650, what is the difference between their raw scores?
    Answer: 300 points First student's raw score: Using z = (x - μ)/σ, we get -1.5 = (x - 500)/100, so x = 500 + (-1.5)(100) = 350. Second student scored 650. The difference is 650 - 350 = 300 points. Choice B results from miscalculating the first student's score as 400. Choice C represents just the z-score times standard deviation (1.5 × 100 + 50). Choice D uses only the z-score portion (1.5 × 100).
13. A standardized test has scores normally distributed with mean 500 and standard deviation 100. Student A scored 650, Student B scored 720, and Student C scored 380. If a scholarship requires applicants to be in the top 2.5% of test takers (z-score ≥ 1.96), which students qualify?
    Answer: Only Student B qualifies with z-score 2.2, while Students A and C have z-scores 1.5 and -1.2 respectively Calculate z-scores: Student A: z = (650-500)/100 = 1.5. Student B: z = (720-500)/100 = 2.2. Student C: z = (380-500)/100 = -1.2. Only Student B has z ≥ 1.96, so only B qualifies. Choice B incorrectly includes Student A whose z-score is below 1.96. Choice C incorrectly calculates Student C's z-score as positive. Choice D incorrectly concludes no one qualifies when Student B clearly does.
14. A psychology professor curves grades so that test scores follow a normal distribution with mean 82 and standard deviation 6. If a student needs a z-score of at least 1.5 to earn an A grade, and another student needs a z-score of at least -0.5 to avoid failing, what is the difference between the minimum raw scores for these two outcomes?
    Answer: 12 points When you encounter normal distribution problems involving z-scores and raw scores, you need to use the z-score formula to convert between standardized and actual values. The z-score formula is $z = \frac{x - \mu}{\sigma}$, where x is the raw score, μ is the mean (82), and σ is the standard deviation (6). Rearranging to solve for x: $x = z\sigma + \mu$. For the A grade (z ≥ 1.5): $x = 1.5(6) + 82 = 9 + 82 = 91$ For avoiding failure (z ≥ -0.5): $x = -0.5(6) + 82 = -3 + 82 = 79$ The difference between these minimum raw scores is $91 - 79 = 12$ points. Let's examine why the other answers are wrong. Choice A (21 points) likely comes from incorrectly calculating the z-score difference as 2.0 and multiplying by standard deviation plus an arithmetic error. Choice B (15 points) might result from using the wrong z-score values or incorrectly applying the formula. Choice C (18 points) could come from multiplying the z-score difference (2.0) by an incorrect standard deviation value or making calculation errors. The key insight is that when you have two z-scores and need the difference in raw scores, you can either calculate each raw score separately (as we did) or recognize that the difference equals the difference in z-scores multiplied by the standard deviation: $(1.5 - (-0.5)) \times 6 = 2.0 \times 6 = 12$.
15. A teacher curves exam scores using z-scores. The original scores are normally distributed with mean 76 and standard deviation 8. If a student scored 84 on the original exam, and the teacher adds 10 points to everyone's score, what is the student's z-score on the curved distribution?
    Answer: z = 1.0, because adding a constant to all scores doesn't change z-scores or standard deviation When you encounter z-score problems involving transformations, the key insight is understanding how different operations affect the mean, standard deviation, and ultimately the z-score itself. A z-score measures how many standard deviations a value is from the mean: $z = \frac{x - \mu}{\sigma}$. Let's trace what happens when the teacher adds 10 points to everyone's score. Originally: mean = 76, standard deviation = 8, student's score = 84 Student's original z-score: $z = \frac{84 - 76}{8} = 1.0$ After adding 10 points to all scores: - New mean: 76 + 10 = 86 - New standard deviation: still 8 (adding a constant doesn't change variability) - Student's new score: 84 + 10 = 94 New z-score: $z = \frac{94 - 86}{8} = \frac{8}{8} = 1.0$ The z-score remains unchanged because adding a constant shifts all values equally, preserving their relative positions. Choice A incorrectly suggests the curve "amplifies" deviation—it doesn't. Choice B makes a calculation error, getting 1.75 instead of 1.0 when computing $\frac{94-86}{8}$. Choice C incorrectly assumes that adding a constant affects the standard deviation (the denominator), which it doesn't—only the relative spread matters. Study tip: Remember that z-scores are unaffected by adding or subtracting constants to all data points. Only operations that change the relative spread (like multiplication) will alter z-scores. This principle appears frequently in standardization problems.
16. A quality control manager knows that the weights of products are normally distributed with a mean of 250 grams and a standard deviation of 18 grams. If a product weighs 232 grams, and the manager wants to find another product that is exactly as far above the mean as this product is below the mean, what should the second product weigh?
    Answer: 268 grams First, find the z-score of the 232-gram product: z = (232 - 250)/18 = -18/18 = -1.0. A product equally far above the mean would have z = +1.0. Converting back: x = 250 + (1.0)(18) = 268 grams. Choice B (286) adds twice the difference to the mean (250 + 36). Choice C (264) represents half the correct distance above the mean. Choice D (272) uses incorrect arithmetic in the conversion.
17. An electronics manufacturer produces resistors with resistance values normally distributed around 100 ohms with standard deviation 5 ohms. A batch contains resistors measuring 92, 103, and 108 ohms. If an engineer needs to select one resistor that represents the most extreme deviation from specification, which should be chosen and why?
    Answer: Either the 92-ohm or 108-ohm resistor because both have z-score magnitude 1.6, equally extreme Calculate z-scores: 92Ω: z = (92-100)/5 = -1.6. 103Ω: z = (103-100)/5 = 0.6. 108Ω: z = (108-100)/5 = 1.6. For most extreme deviation, we want largest |z|. Both 92Ω and 108Ω have |z| = 1.6, making them equally extreme. Choice A and B each identify one extreme but miss that they're equally extreme. Choice D misunderstands the question's requirement.
18. In a manufacturing process, the weights of widgets are normally distributed with mean 250 grams and standard deviation 15 grams. Two widgets are randomly selected: Widget A weighs 280 grams and Widget B weighs 265 grams. A quality control inspector claims that Widget A is "more unusual" than Widget B. Which statement best evaluates this claim?
    Answer: The claim is correct because Widget A has a z-score of 2.0 while Widget B has a z-score of 1.0, and |2.0| > |1.0| To determine which widget is more unusual, we calculate z-scores. For Widget A: z = (280-250)/15 = 30/15 = 2.0. For Widget B: z = (265-250)/15 = 15/15 = 1.0. Since |2.0| > |1.0|, Widget A is indeed more unusual. Choice B has incorrect z-score calculations. Choice C incorrectly uses raw weights instead of standardized scores. Choice D incorrectly states both have the same z-score magnitude.
19. A factory produces items with weights normally distributed around 50 grams with standard deviation 8 grams. An inspector finds three items with weights 42g, 58g, and 66g. If the inspector can only investigate one item due to time constraints, which should be prioritized based on how unusual it is?
    Answer: The 66g item should be investigated because it has z-score 2.0, the most extreme deviation Calculate z-scores: 42g: z = (42-50)/8 = -1.0. 58g: z = (58-50)/8 = 1.0. 66g: z = (66-50)/8 = 2.0. The most unusual item has the largest absolute z-score, which is the 66g item with |2.0| = 2.0. Choice A correctly calculates but focuses on direction rather than magnitude. Choice C correctly calculates but chooses a less extreme item. Choice D misses the point that among acceptable items, some are still more unusual.
20. A quality control manager monitors production times that are normally distributed with mean 45 minutes and standard deviation 6 minutes. Three products have completion times of 39, 48, and 54 minutes. If the manager wants to identify which product's completion time is most typical, which should be selected?
    Answer: The 48-minute product because its z-score of 0.5 is closest to zero, making it most typical Calculate z-scores: 39 min: z = (39-45)/6 = -1.0. 48 min: z = (48-45)/6 = 0.5. 54 min: z = (54-45)/6 = 1.5. The most typical value has z-score closest to 0, which is the 48-minute product with z = 0.5. Choice A correctly calculates but misinterprets what 'typical' means. Choice C correctly calculates but chooses a less typical value. Choice D uses raw differences instead of standardized scores.

Start this quiz