Z-Scores & Standardization - College Statistics

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?
Answer: 14.0 Let $\mu$ be the mean and $\sigma$ be the standard deviation. We can set up a system of two equations: 1) $1.5 = (85 - \mu) / \sigma \Rightarrow 1.5\sigma = 85 - \mu$ 2) $-0.75 = (64 - \mu) / \sigma \Rightarrow -0.75\sigma = 64 - \mu$ To solve for $\sigma$ , we can subtract the second equation from the first: $(1.5\sigma) - (-0.75\sigma) = (85 - \mu) - (64 - \mu)$ . This simplifies to $2.25\sigma = 85 - 64$ . $2.25\sigma = 21$ . $\sigma = 21 / 2.25 = 14$ . The standard deviation is 14.0.
If the z-score for a data value $x$ is $z_1$ with respect to a dataset with mean $\mu$ and standard deviation $\sigma$ , what would be the new z-score for $x$ if the mean of the dataset increased by $\sigma/2$ while the standard deviation remained unchanged?
Answer: $z_1 - 0.5$ When you encounter z-score transformation problems, focus on how changes to the dataset parameters (mean and standard deviation) affect the standardization formula: $z = \frac{x - \mu}{\sigma}$ . Let's work through this step-by-step. Initially, you have $z_1 = \frac{x - \mu}{\sigma}$ . When the mean increases by $\sigma/2$ , the new mean becomes $\mu + \sigma/2$ , while the standard deviation remains $\sigma$ . The new z-score is: $z_{new} = \frac{x - (\mu + \sigma/2)}{\sigma} = \frac{x - \mu - \sigma/2}{\sigma} = \frac{x - \mu}{\sigma} - \frac{\sigma/2}{\sigma} = z_1 - 0.5$ This makes intuitive sense: when the dataset mean increases but your data point $x$ stays the same, that point becomes relatively smaller compared to the new center, so its z-score decreases. Looking at the wrong answers: Choice A ( $2z_1$ ) incorrectly suggests the z-score doubles, ignoring that only the mean changed. Choice B ( $z_1 + 0.5$ ) gets the magnitude right but the wrong sign—this would be correct if the mean had decreased. Choice C ( $z_1/2$ ) incorrectly assumes the z-score is halved, which would happen if the standard deviation doubled, not if the mean changed. Study tip: Remember that z-scores measure how many standard deviations a point is from the mean. When the mean shifts toward your data point, the z-score decreases; when it shifts away, the z-score increases. Always track the direction of change carefully.
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?
Answer: The sum of the z-scores will be exactly 0. 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 $\sum z_i = \sum \frac{x_i - \mu}{\sigma} = \frac{1}{\sigma} \sum(x_i - \mu)$ . A fundamental property of the mean is that the sum of the deviations from the mean, $\sum(x_i - \mu)$ , 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.
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 $Y = 100 - 2X$ . What is the standardized score (z-score) for the 30°C reading on the new Y scale?
Answer: -2.0 First, calculate the original z-score for X = 30°C: $z_X = (30 - 20) / 5 = 10 / 5 = 2.0$ . Next, determine the parameters of the new scale Y. The new mean is $\mu_Y = 100 - 2\mu_X = 100 - 2(20) = 60$ . The new standard deviation is $\sigma_Y = |-2|\sigma_X = 2(5) = 10$ . The new value on the Y scale is $Y = 100 - 2(30) = 40$ . The new z-score is $z_Y = (40 - 60) / 10 = -20 / 10 = -2.0$ . Note that for a linear transformation $Y = aX+b$ , the new z-score is equal to the old z-score if $a>0$ and is the negative of the old z-score if $a<0$ .
Two data points from a distribution with a mean of 50 are 40 and 60. Which of the following statements about their z-scores is true?
Answer: Their z-scores are equal in magnitude but opposite in sign. Let the standard deviation be $\sigma$ . The z-score for 40 is $z_{40} = (40 - 50) / \sigma = -10 / \sigma$ . The z-score for 60 is $z_{60} = (60 - 50) / \sigma = 10 / \sigma$ . Since the two data points are equidistant from the mean (both are 10 units away), their z-scores have the same magnitude ( $|-10/\sigma| = |10/\sigma|$ ) but opposite signs. Therefore, they are equal in magnitude but opposite in sign.
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?
Answer: It will become negative because the combined mean is higher than her score. 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: $\mu_{comb} = (30 \times 70 + 20 \times 85) / (30 + 20) = (2100 + 1700) / 50 = 3800 / 50 = 76$ . Since the student's score of 70 is less than the new combined mean of 76, the numerator of her z-score calculation $(x - \mu_{comb})$ will be negative ( $70 - 76 = -6$ ). Therefore, her z-score in the combined group must be negative.
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?
Answer: Performance was stronger on the physics exam, as indicated by a z-score of 2.4. To compare scores from different distributions, we must standardize them by calculating their z-scores. Biology z-score: $z = (88 - 76) / 8 = 12 / 8 = 1.5$ . Physics z-score: $z = (82 - 70) / 5 = 12 / 5 = 2.4$ . 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.
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?
Answer: 75 The problem describes the calculation of a z-score where the mean ( $\mu$ ) is 50 and the standard deviation ( $\sigma$ ) is 10. The formula is $z = (x - \mu) / \sigma$ . We are given $z = 2.5$ , $\mu = 50$ , and $\sigma = 10$ , and we need to find the original value, $x$ . $2.5 = (x - 50) / 10$ . To solve for $x$ , first multiply both sides by 10: $2.5 \times 10 = x - 50$ , which simplifies to $25 = x - 50$ . Then, add 50 to both sides: $x = 25 + 50 = 75$ . The original value was 75.
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?
Answer: 1.5 The z-score is invariant under a positive linear transformation ( $Y = aX + b$ where $a > 0$ ). We can show this by calculation. The original z-score for a score of 75 is $z_{old} = (75 - 60) / 10 = 1.5$ . The new score is $Y = 1.2(75) + 5 = 90 + 5 = 95$ . The new mean is $\mu_Y = 1.2(60) + 5 = 72 + 5 = 77$ . The new standard deviation is $\sigma_Y = 1.2(10) = 12$ . The new z-score is $z_{new} = (95 - 77) / 12 = 18 / 12 = 1.5$ . The z-score has not changed.
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?
Answer: It increased by 0.25 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): $z_{initial} = (70 - 60) / 8 = 10 / 8 = 1.25$ . Final z-score (at 72 lbs): $z_{final} = (72 - 60) / 8 = 12 / 8 = 1.50$ . The change in the z-score is $z_{final} - z_{initial} = 1.50 - 1.25 = 0.25$ . The z-score increased by 0.25.
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?
Answer: Strongly skewed to the right, identical in shape to the original distribution. 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.
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?
Answer: 0, regardless of the distribution's shape. 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: $z = (x - \mu) / \sigma$ . If we substitute the mean for $x$ , we get $z = (\mu - \mu) / \sigma = 0 / \sigma = 0$ . This is true for any distribution, regardless of its shape.
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?
Answer: 1 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 $1^2 = 1$ . This property is always true regardless of the original dataset's characteristics or size, as long as the original standard deviation is not zero.
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?
Answer: -1.50 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: $z = \frac{x - \mu}{\sigma}$ , 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 $\sigma = \sqrt{16} = 4$ . Now you can calculate: $z = \frac{44 - 50}{4} = \frac{-6}{4} = -1.50$ . 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: $\frac{-6}{16} = -0.375$ . 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.
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?
Answer: 5 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 $\sigma_z^2 = \frac{\sum (z_i - \mu_z)^2}{N}$ . Since $\sigma_z^2 = 1$ and $\mu_z = 0$ , the formula becomes $1 = \frac{\sum z_i^2}{N}$ . We are asked for the sum of the squares of the z-scores, which is $\sum z_i^2$ . From the formula, $\sum z_i^2 = N \times 1 = N$ . In this case, there are N=5 data points, so the sum of the squares of the z-scores is 5.
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?
Answer: The bolt from Process A is more unusual, with a z-score of -2.5. 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: $z_A = (4.95 - 5.00) / 0.02 = -0.05 / 0.02 = -2.5$ . Z-score for Bolt B: $z_B = (5.22 - 5.10) / 0.05 = 0.12 / 0.05 = 2.4$ . Comparing the absolute values of the z-scores, $|-2.5| > |2.4|$ . Therefore, the bolt from Process A is more unusual relative to its process.
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?
Answer: The stock's return was 1.5 standard deviations below the average historical return. 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.
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?
Answer: 75 The formula for a z-score is $z = (x - \mu) / \sigma$ . We are given $z = -0.8$ , $x = 450$ , and $\mu = 510$ . We need to solve for $\sigma$ . Plugging in the values: $-0.8 = (450 - 510) / \sigma$ . $-0.8 = -60 / \sigma$ . Multiplying both sides by $\sigma$ gives $-0.8\sigma = -60$ . Dividing by -0.8 gives $\sigma = -60 / -0.8 = 75$ .
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?
Answer: It will increase. The original z-score for 115 is $(115 - 100) / 15 = 1.0$ . The removed value, 130, is above the mean. Removing a value that is above the mean will cause the new mean ( $\mu_{new}$ ) to decrease, so $\mu_{new} < 100$ . The removed value is also far from the mean, so its removal will cause the new standard deviation ( $\sigma_{new}$ ) to decrease, so $\sigma_{new} < 15$ . The new z-score for 115 will be $z_{new} = (115 - \mu_{new}) / \sigma_{new}$ . Since $\mu_{new} < 100$ , the numerator $(115 - \mu_{new})$ will be greater than 15. Since $\sigma_{new} < 15$ , 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.
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 ( $\mu$ ) of the dataset?
Answer: 90 Let $\mu$ be the mean and $\sigma$ be the standard deviation. The z-score formula is $z = (x - \mu) / \sigma$ . We can set up a system of two linear equations: 1) $2.0 = (110 - \mu) / \sigma \Rightarrow 2\sigma = 110 - \mu$ 2) $-1.0 = (80 - \mu) / \sigma \Rightarrow -\sigma = 80 - \mu$ From the second equation, we can write $\sigma = \mu - 80$ . Substituting this into the first equation gives: $2(\mu - 80) = 110 - \mu$ . Distributing gives $2\mu - 160 = 110 - \mu$ . Solving for $\mu$ , we get $3\mu = 270$ , so $\mu = 90$ .

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