College Statistics Quiz: Z Scores And Standardization
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Z Scores And StandardizationQuestion 1 of 20

A dataset consisting of 25 observations has a mean of 50 and a variance of 16. What would be the z-score for an observation of 44?

-6.00
-0.375
-0.25
-1.50
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College Statistics Quiz

College Statistics Quiz: Z Scores And Standardization

Practice Z Scores And Standardization in College Statistics 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 Z Scores And Standardization, giving you a quick way to practice the rules, question types, and explanations that matter most for College Statistics.

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 dataset consisting of 25 observations has a mean of 50 and a variance of 16. What would be the z-score for an observation of 44?

  1. -6.00
  2. -0.375
  3. -0.25
  4. -1.50 (correct answer)
Explanation: When you encounter z-score problems, you're working with standardized values that tell you how many standard deviations an observation is from the mean. The z-score formula is: z=x−μσz = \frac{x - \mu}{\sigma}, where x is your observation, μ is the mean, and σ is the standard deviation. Here, you have an observation of 44, a mean of 50, and a variance of 16. Since variance equals the standard deviation squared, you need to find σ=16=4\sigma = \sqrt{16} = 4. Now you can calculate: z=44−504=−64=−1.50z = \frac{44 - 50}{4} = \frac{-6}{4} = -1.50. The negative value indicates the observation is below the mean. Looking at the wrong answers: Choice A (-6.00) represents using the numerator alone without dividing by the standard deviation—this ignores the scaling aspect of standardization. Choice B (-0.375) appears to result from incorrectly using the variance (16) as the denominator instead of the standard deviation: −616=−0.375\frac{-6}{16} = -0.375. Choice C (-0.25) likely comes from dividing by 24 (perhaps confusing this with degrees of freedom) rather than the standard deviation. The correct answer is D (-1.50), meaning this observation falls 1.5 standard deviations below the mean. Remember: always convert variance to standard deviation by taking the square root before calculating z-scores. The most common mistake is forgetting this conversion step and using variance directly in the denominator.

Question 2

A dataset consisting of 25 observations has a mean of 50 and a variance of 16. What would be the z-score for an observation of 44?

  1. -6.00
  2. -0.375
  3. -0.25
  4. -1.50 (correct answer)
Explanation: When you encounter z-score problems, you're working with standardized values that tell you how many standard deviations an observation is from the mean. The z-score formula is: z=x−μσz = \frac{x - \mu}{\sigma}, where x is your observation, μ is the mean, and σ is the standard deviation. Here, you have an observation of 44, a mean of 50, and a variance of 16. Since variance equals the standard deviation squared, you need to find σ=16=4\sigma = \sqrt{16} = 4. Now you can calculate: z=44−504=−64=−1.50z = \frac{44 - 50}{4} = \frac{-6}{4} = -1.50. The negative value indicates the observation is below the mean. Looking at the wrong answers: Choice A (-6.00) represents using the numerator alone without dividing by the standard deviation—this ignores the scaling aspect of standardization. Choice B (-0.375) appears to result from incorrectly using the variance (16) as the denominator instead of the standard deviation: −616=−0.375\frac{-6}{16} = -0.375. Choice C (-0.25) likely comes from dividing by 24 (perhaps confusing this with degrees of freedom) rather than the standard deviation. The correct answer is D (-1.50), meaning this observation falls 1.5 standard deviations below the mean. Remember: always convert variance to standard deviation by taking the square root before calculating z-scores. The most common mistake is forgetting this conversion step and using variance directly in the denominator.

Question 3

A company evaluates two manufacturing processes. Process A produces bolts with a mean length of 5.00 cm and a standard deviation of 0.02 cm. Process B produces bolts with a mean length of 5.10 cm and a standard deviation of 0.05 cm. A bolt from Process A is measured at 4.95 cm. A bolt from Process B is measured at 5.22 cm. Which statement correctly assesses the 'unusualness' of these two bolts relative to their respective processes?

  1. The bolt from Process A is more unusual, with a z-score of -2.5. (correct answer)
  2. The bolt from Process B is more unusual, with a z-score of 2.4.
  3. The bolt from Process B is more unusual because its length differs from its mean by 0.12 cm, which is greater than 0.05 cm for the bolt from Process A.
  4. Both bolts are similarly unusual as their z-scores are approximately equal in magnitude.
Explanation: To compare unusualness, we calculate the z-score for each bolt. A larger absolute z-score indicates a more unusual value. Z-score for Bolt A: zA=(4.95−5.00)/0.02=−0.05/0.02=−2.5z_A = (4.95 - 5.00) / 0.02 = -0.05 / 0.02 = -2.5. Z-score for Bolt B: zB=(5.22−5.10)/0.05=0.12/0.05=2.4z_B = (5.22 - 5.10) / 0.05 = 0.12 / 0.05 = 2.4. Comparing the absolute values of the z-scores, ∣−2.5∣>∣2.4∣|-2.5| > |2.4|. Therefore, the bolt from Process A is more unusual relative to its process.

Question 4

A set of temperature readings in Celsius (X) has a mean of 20 and a standard deviation of 5. One reading is 30°C. These temperatures are then converted to a new scale, Y, using the formula Y=100−2XY = 100 - 2X. What is the standardized score (z-score) for the 30°C reading on the new Y scale?

  1. -2.0 (correct answer)
  2. 2.0
  3. -4.0
  4. 4.0
Explanation: First, calculate the original z-score for X = 30°C: zX=(30−20)/5=10/5=2.0z_X = (30 - 20) / 5 = 10 / 5 = 2.0. Next, determine the parameters of the new scale Y. The new mean is μY=100−2μX=100−2(20)=60\mu_Y = 100 - 2\mu_X = 100 - 2(20) = 60. The new standard deviation is σY=∣−2∣σX=2(5)=10\sigma_Y = |-2|\sigma_X = 2(5) = 10. The new value on the Y scale is Y=100−2(30)=40Y = 100 - 2(30) = 40. The new z-score is zY=(40−60)/10=−20/10=−2.0z_Y = (40 - 60) / 10 = -20 / 10 = -2.0. Note that for a linear transformation Y=aX+bY = aX+b, the new z-score is equal to the old z-score if a>0a>0 and is the negative of the old z-score if a<0a<0.

Question 5

The distribution of weights for a certain breed of dog is approximately symmetric with a mean of 60 pounds and a standard deviation of 8 pounds. A particular dog of this breed weighs 70 pounds. The dog's owner changes its diet, and after a month, the dog weighs 72 pounds. Assuming the mean and standard deviation for the breed remain unchanged, how did the dog's z-score change?

  1. It increased by 0.25 (correct answer)
  2. It increased by 2.00
  3. It is now 1.50
  4. It is now 9.00
Explanation: This is a multi-step problem. First, calculate the initial z-score, then the final z-score, and finally find the difference. Initial z-score (at 70 lbs): zinitial=(70−60)/8=10/8=1.25z_{initial} = (70 - 60) / 8 = 10 / 8 = 1.25. Final z-score (at 72 lbs): zfinal=(72−60)/8=12/8=1.50z_{final} = (72 - 60) / 8 = 12 / 8 = 1.50. The change in the z-score is zfinal−zinitial=1.50−1.25=0.25z_{final} - z_{initial} = 1.50 - 1.25 = 0.25. The z-score increased by 0.25.

Question 6

A dataset of 50 exam scores is converted into a set of z-scores. Which of the following statements about the resulting set of z-scores is guaranteed to be true, regardless of the shape of the original distribution of scores?

  1. The distribution of z-scores will be approximately normal.
  2. The median of the z-scores will be exactly 0.
  3. The sum of the z-scores will be exactly 0. (correct answer)
  4. The range of the z-scores will be from -3 to 3.
Explanation: Standardizing a dataset converts it to a new distribution with a mean of 0 and a standard deviation of 1. The sum of the z-scores is given by ∑zi=∑xi−μσ=1σ∑(xi−μ)\sum z_i = \sum \frac{x_i - \mu}{\sigma} = \frac{1}{\sigma} \sum(x_i - \mu). A fundamental property of the mean is that the sum of the deviations from the mean, ∑(xi−μ)\sum(x_i - \mu), is always 0. Therefore, the sum of the z-scores is also guaranteed to be 0. Standardization does not change the shape of the distribution, so A is incorrect. The median is only guaranteed to be 0 if the original distribution is symmetric, so B is incorrect. The range is not guaranteed to be within [-3, 3], so D is incorrect.

Question 7

A dataset has a mean of 100 and a standard deviation of 15. A data point with a value of 130 is removed from the dataset. Consider a data point with a value of 115 that remains in the dataset. How will its z-score change after the removal of the value 130?

  1. It will increase. (correct answer)
  2. It will decrease.
  3. It will remain exactly the same.
  4. The direction of change cannot be determined without knowing the sample size.
Explanation: The original z-score for 115 is (115−100)/15=1.0(115 - 100) / 15 = 1.0. The removed value, 130, is above the mean. Removing a value that is above the mean will cause the new mean (μnew\mu_{new}) to decrease, so μnew<100\mu_{new} < 100. The removed value is also far from the mean, so its removal will cause the new standard deviation (σnew\sigma_{new}) to decrease, so σnew<15\sigma_{new} < 15. The new z-score for 115 will be znew=(115−μnew)/σnewz_{new} = (115 - \mu_{new}) / \sigma_{new}. Since μnew<100\mu_{new} < 100, the numerator (115−μnew)(115 - \mu_{new}) will be greater than 15. Since σnew<15\sigma_{new} < 15, the denominator will be smaller than 15. Dividing a larger numerator by a smaller denominator results in a larger value, so the z-score for 115 will increase.

Question 8

The distribution of weights for a certain breed of dog is approximately symmetric with a mean of 60 pounds and a standard deviation of 8 pounds. A particular dog of this breed weighs 70 pounds. The dog's owner changes its diet, and after a month, the dog weighs 72 pounds. Assuming the mean and standard deviation for the breed remain unchanged, how did the dog's z-score change?

  1. It increased by 0.25 (correct answer)
  2. It increased by 2.00
  3. It is now 1.50
  4. It is now 9.00
Explanation: This is a multi-step problem. First, calculate the initial z-score, then the final z-score, and finally find the difference. Initial z-score (at 70 lbs): zinitial=(70−60)/8=10/8=1.25z_{initial} = (70 - 60) / 8 = 10 / 8 = 1.25. Final z-score (at 72 lbs): zfinal=(72−60)/8=12/8=1.50z_{final} = (72 - 60) / 8 = 12 / 8 = 1.50. The change in the z-score is zfinal−zinitial=1.50−1.25=0.25z_{final} - z_{initial} = 1.50 - 1.25 = 0.25. The z-score increased by 0.25.

Question 9

A professor grades an exam and finds the scores have a mean of 60 and a standard deviation of 10. To adjust the grades, she applies a linear transformation to every score: first, she multiplies each score by 1.2, and then she adds 5 points. What is the new z-score for a student who originally scored a 75?

  1. 1.5 (correct answer)
  2. 1.8
  3. 3.5
  4. 6.8
Explanation: The z-score is invariant under a positive linear transformation (Y=aX+bY = aX + b where a>0a > 0). We can show this by calculation. The original z-score for a score of 75 is zold=(75−60)/10=1.5z_{old} = (75 - 60) / 10 = 1.5. The new score is Y=1.2(75)+5=90+5=95Y = 1.2(75) + 5 = 90 + 5 = 95. The new mean is μY=1.2(60)+5=72+5=77\mu_Y = 1.2(60) + 5 = 72 + 5 = 77. The new standard deviation is σY=1.2(10)=12\sigma_Y = 1.2(10) = 12. The new z-score is znew=(95−77)/12=18/12=1.5z_{new} = (95 - 77) / 12 = 18 / 12 = 1.5. The z-score has not changed.

Question 10

A large dataset representing annual income in a metropolitan area is known to be strongly skewed to the right. If every income value in this dataset is converted to its corresponding z-score, what will be the shape of the resulting distribution of z-scores?

  1. Approximately normal, because standardization is a normalizing transformation.
  2. Symmetrical but not necessarily normal, with a mean of 0.
  3. Strongly skewed to the right, identical in shape to the original distribution. (correct answer)
  4. Uniform, because the data points are rescaled to be evenly distributed around 0.
Explanation: Standardization is a linear transformation. It changes the center (mean becomes 0) and the scale (standard deviation becomes 1) of the data, but it does not change the shape of the distribution. Properties like skewness, kurtosis, and modality remain unchanged. Therefore, if the original distribution of incomes is strongly skewed to the right, the distribution of z-scores will also be strongly skewed to the right.

Question 11

A student's score on a test is 85, and it corresponds to a z-score of 1.5. Another student's score of 64 on the same test corresponds to a z-score of -0.75. What is the standard deviation of the test scores?

  1. 7.0
  2. 10.5
  3. 14.0 (correct answer)
  4. 21.0
Explanation: Let μ\mu be the mean and σ\sigma be the standard deviation. We can set up a system of two equations:
  1. 1.5=(85−μ)/σ⇒1.5σ=85−μ1.5 = (85 - \mu) / \sigma \Rightarrow 1.5\sigma = 85 - \mu
  2. −0.75=(64−μ)/σ⇒−0.75σ=64−μ-0.75 = (64 - \mu) / \sigma \Rightarrow -0.75\sigma = 64 - \mu To solve for σ\sigma, we can subtract the second equation from the first: (1.5σ)−(−0.75σ)=(85−μ)−(64−μ)(1.5\sigma) - (-0.75\sigma) = (85 - \mu) - (64 - \mu). This simplifies to 2.25σ=85−642.25\sigma = 85 - 64. 2.25σ=212.25\sigma = 21. σ=21/2.25=14\sigma = 21 / 2.25 = 14. The standard deviation is 14.0.

Question 12

Class A (30 students) has an exam mean of 70 and a standard deviation of 10. Class B (20 students) has an exam mean of 85 and a standard deviation of 5. The two classes are combined to form one group of 50 students. A student from Class A who scored 70 on the exam had a z-score of 0 within her class. Which statement best describes her z-score in the combined group?

  1. It will remain 0 because her score was equal to her original class mean.
  2. It will become positive because the overall spread of scores is likely larger.
  3. It will become negative because the combined mean is higher than her score. (correct answer)
  4. The sign of the z-score cannot be determined without the combined standard deviation.
Explanation: The student's score is 70. To find her new z-score, we must first find the mean of the combined group. The combined mean is the weighted average of the individual class means: μcomb=(30×70+20×85)/(30+20)=(2100+1700)/50=3800/50=76\mu_{comb} = (30 \times 70 + 20 \times 85) / (30 + 20) = (2100 + 1700) / 50 = 3800 / 50 = 76. Since the student's score of 70 is less than the new combined mean of 76, the numerator of her z-score calculation (x−μcomb)(x - \mu_{comb}) will be negative (70−76=−670 - 76 = -6). Therefore, her z-score in the combined group must be negative.

Question 13

In a dataset, a value of 110 has a z-score of 2.0, and a value of 80 has a z-score of -1.0. What is the mean (μ\mu) of the dataset?

  1. 85
  2. 90 (correct answer)
  3. 95
  4. 100
Explanation: Let μ\mu be the mean and σ\sigma be the standard deviation. The z-score formula is z=(x−μ)/σz = (x - \mu) / \sigma. We can set up a system of two linear equations:
  1. 2.0=(110−μ)/σ⇒2σ=110−μ2.0 = (110 - \mu) / \sigma \Rightarrow 2\sigma = 110 - \mu
  2. −1.0=(80−μ)/σ⇒−σ=80−μ-1.0 = (80 - \mu) / \sigma \Rightarrow -\sigma = 80 - \mu From the second equation, we can write σ=μ−80\sigma = \mu - 80. Substituting this into the first equation gives: 2(μ−80)=110−μ2(\mu - 80) = 110 - \mu. Distributing gives 2μ−160=110−μ2\mu - 160 = 110 - \mu. Solving for μ\mu, we get 3μ=2703\mu = 270, so μ=90\mu = 90.

Question 14

A student's score on a national biology exam was 88, where the national mean was 76 and the standard deviation was 8. The same student's score on a local physics exam was 82, where the local mean was 70 and the standard deviation was 5. Which statement accurately compares the student's performance on the two exams?

  1. Performance was stronger on the physics exam, as indicated by a z-score of 2.4. (correct answer)
  2. Performance was stronger on the biology exam, as indicated by a z-score of 1.5.
  3. Performance was stronger on the biology exam because the raw score of 88 is higher than 82.
  4. Performance was equal on both because the score was 12 points above the mean on both exams.
Explanation: To compare scores from different distributions, we must standardize them by calculating their z-scores. Biology z-score: z=(88−76)/8=12/8=1.5z = (88 - 76) / 8 = 12 / 8 = 1.5. Physics z-score: z=(82−70)/5=12/5=2.4z = (82 - 70) / 5 = 12 / 5 = 2.4. Since the z-score for the physics exam (2.4) is higher than the z-score for the biology exam (1.5), the student's performance was stronger relative to their peers on the physics exam.

Question 15

In a distribution of scores for a final exam, a student's score of 450 corresponded to a z-score of -0.8. The mean score for the exam was 510. What was the standard deviation of the exam scores?

  1. 48
  2. 60
  3. 75 (correct answer)
  4. 93.75
Explanation: The formula for a z-score is z=(x−μ)/σz = (x - \mu) / \sigma. We are given z=−0.8z = -0.8, x=450x = 450, and μ=510\mu = 510. We need to solve for σ\sigma. Plugging in the values: −0.8=(450−510)/σ-0.8 = (450 - 510) / \sigma. −0.8=−60/σ-0.8 = -60 / \sigma. Multiplying both sides by σ\sigma gives −0.8σ=−60-0.8\sigma = -60. Dividing by -0.8 gives σ=−60/−0.8=75\sigma = -60 / -0.8 = 75.

Question 16

For any dataset with a non-zero standard deviation, the entire dataset is converted into z-scores. What is the variance of this new set of z-scores?

  1. 0
  2. 1 (correct answer)
  3. Equal to the original variance of the dataset.
  4. Dependent on the number of data points in the dataset.
Explanation: By definition, standardizing a dataset transforms the data to have a mean of 0 and a standard deviation of 1. Variance is the square of the standard deviation. Therefore, the variance of the set of z-scores will be 12=11^2 = 1. This property is always true regardless of the original dataset's characteristics or size, as long as the original standard deviation is not zero.

Question 17

A quality control inspector measures the weight of 100 widgets produced by a machine. She calculates the mean weight and the standard deviation for this sample. What is the z-score corresponding to the mean weight of the sample?

  1. 0, but only if the distribution of weights is symmetric.
  2. 1, because the mean is one standard unit.
  3. 0, regardless of the distribution's shape. (correct answer)
  4. It cannot be determined without the value of the standard deviation.
Explanation: The z-score measures how many standard deviations a data point is from the mean. The mean itself is 0 standard deviations from the mean. This can be seen from the formula: z=(x−μ)/σz = (x - \mu) / \sigma. If we substitute the mean for xx, we get z=(μ−μ)/σ=0/σ=0z = (\mu - \mu) / \sigma = 0 / \sigma = 0. This is true for any distribution, regardless of its shape.

Question 18

An investment analyst states that a particular stock's daily return had a z-score of -1.5 yesterday, based on the historical distribution of daily returns. Which is the most accurate interpretation of this statement?

  1. The stock's return was 1.5 times lower than the average historical return.
  2. The stock's return was 1.5 standard deviations below the average historical return. (correct answer)
  3. The stock's return was negative, and its magnitude was 1.5% of the average return.
  4. The probability of observing a return this low or lower is approximately 1.5%.
Explanation: A z-score represents the number of standard deviations a data point is from the mean of its distribution. A positive z-score indicates the value is above the mean, while a negative z-score indicates it is below the mean. A z-score of -1.5 means the data point (the stock's return) is exactly 1.5 standard deviations below the mean of the historical returns.

Question 19

A dataset contains the following five numbers: 2, 3, 5, 8, 12. The mean of this dataset is 6.0 and the standard deviation is approximately 3.61. If a new dataset is created by converting each of these five numbers to its z-score, what will be the sum of the squares of the values in the new dataset?

  1. 1
  2. 5 (correct answer)
  3. 13.0
  4. 25
Explanation: When a dataset is standardized, the resulting set of z-scores has a mean of 0 and a standard deviation of 1. The variance, which is the square of the standard deviation, is therefore also 1. The formula for the population variance of the z-scores is σz2=∑(zi−μz)2N\sigma_z^2 = \frac{\sum (z_i - \mu_z)^2}{N}. Since σz2=1\sigma_z^2 = 1 and μz=0\mu_z = 0, the formula becomes 1=∑zi2N1 = \frac{\sum z_i^2}{N}. We are asked for the sum of the squares of the z-scores, which is ∑zi2\sum z_i^2. From the formula, ∑zi2=N×1=N\sum z_i^2 = N \times 1 = N. In this case, there are N=5 data points, so the sum of the squares of the z-scores is 5.

Question 20

A statistician standardizes a dataset by subtracting 50 from each value and then dividing by 10. The resulting z-score for a particular value is 2.5. What was the original value?

  1. 52.5
  2. 60
  3. 75 (correct answer)
  4. 125
Explanation: The problem describes the calculation of a z-score where the mean (μ\mu) is 50 and the standard deviation (σ\sigma) is 10. The formula is z=(x−μ)/σz = (x - \mu) / \sigma. We are given z=2.5z = 2.5, μ=50\mu = 50, and σ=10\sigma = 10, and we need to find the original value, xx. 2.5=(x−50)/102.5 = (x - 50) / 10. To solve for xx, first multiply both sides by 10: 2.5×10=x−502.5 \times 10 = x - 50, which simplifies to 25=x−5025 = x - 50. Then, add 50 to both sides: x=25+50=75x = 25 + 50 = 75. The original value was 75.