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Measures of Spread - College Statistics

  1. A dataset of exam scores has a standard deviation of 12 points. The professor decides to curve the exam by multiplying every student's score by 1.1 and then adding 5 points. What will be the new standard deviation of the scores, rounded to the nearest tenth?
    Answer: 13.2 points Measures of spread, such as the standard deviation, are affected by multiplication or division, but not by addition or subtraction. Adding 5 points to every score shifts the entire distribution but does not change its spread. Multiplying each score by 1.1 will multiply the standard deviation by 1.1. The new standard deviation will be 12×1.1=13.212 \times 1.1 = 13.2.
  2. Consider three datasets, each with 100 data points and a range from 0 to 10. Dataset X: A uniform distribution of points spread evenly from 0 to 10. Dataset Y: A U-shaped distribution, with 50 points at the value 0 and 50 points at the value 10. Dataset Z: A single-value distribution, with all 100 points at the value 5. Which of the following correctly orders the datasets by their standard deviation, from smallest to largest?
    Answer: Z < X < Y Standard deviation measures spread from the mean. - Dataset Z has all points identical (at the mean of 5), so there is no spread. Its standard deviation is exactly 0. This is the smallest. - Dataset Y has all its points at the maximum possible distance from the mean (which is 5). This represents the maximum possible spread, so it will have the largest standard deviation. - Dataset X has its points spread out evenly. Its spread is greater than Z's but less concentrated at the extremes than Y's. Therefore, the correct order of standard deviations from smallest to largest is Z, then X, then Y.
  3. For a dataset that is strongly skewed to the right and contains several high-value outliers, which of the following is the most likely relationship between its measures of spread?
    Answer: The standard deviation is significantly larger than the interquartile range. The standard deviation is calculated using the mean and is sensitive to every data point. Outliers and strong skewness pull the mean towards the tail and dramatically increase the squared deviations, thus inflating the standard deviation. The interquartile range (IQR) is based on the 25th and 75th percentiles (quartiles), which are resistant to the effects of outliers. Therefore, in a dataset with high-value outliers and right skew, the standard deviation will be much larger than the more robust IQR. The range can never be smaller than the standard deviation.
  4. A class is split into two sections for an exam. Section A has 10 students with a mean score of 80 and a score variance of 25. Section B has 20 students with a mean score of 80 and a score variance of 49. If the scores from the two sections are combined into a single dataset of 30 students, what is the variance of the combined scores?
    Answer: 41.0 When two groups with the same mean are combined, the variance of the combined group is the weighted average of the individual variances, where the weights are the group sizes. Let nA,nBn_A, n_B be the sizes and σA2,σB2\sigma_A^2, \sigma_B^2 be the variances. The combined variance σC2\sigma_C^2 is given by σC2=nAσA2+nBσB2nA+nB\sigma_C^2 = \frac{n_A \sigma_A^2 + n_B \sigma_B^2}{n_A + n_B}. Plugging in the values: σC2=(10)(25)+(20)(49)10+20=250+98030=123030=41.0\sigma_C^2 = \frac{(10)(25) + (20)(49)}{10 + 20} = \frac{250 + 980}{30} = \frac{1230}{30} = 41.0.
  5. Consider the dataset {2, 5, 8, 11, 14}. Which of the following single-value changes would result in the largest increase in the standard deviation of the dataset?
    Answer: Changing the 14 to a 20 The standard deviation measures the spread of data around the mean. The original mean is 8. The standard deviation is heavily influenced by points far from the mean because it is based on squared deviations. Changing an extreme value (like 14) to be even more extreme (20) will cause the largest increase in a squared deviation term, and thus the largest increase in the standard deviation. Changing a central value (like 8) has a minimal effect. Changing 2 to 0 is a change of 2 units, while changing 14 to 20 is a change of 6 units, making the latter far more impactful.
  6. A dataset has a first quartile (Q1) of 60 and a third quartile (Q3) of 90. According to the 1.5 x IQR rule for identifying potential outliers, which of the following data points would be classified as an outlier?
    Answer: 140 First, calculate the interquartile range (IQR): IQR = Q3 - Q1 = 90 - 60 = 30. Next, determine the fences for outliers. The lower fence is Q1 - 1.5 IQR = 60 - 1.5 30 = 60 - 45 = 15. The upper fence is Q3 + 1.5 IQR = 90 + 1.5 30 = 90 + 45 = 135. Any data point below the lower fence (< 15) or above the upper fence (> 135) is considered an outlier. The value 140 is greater than 135, so it is an outlier. The values 15 and 135 are exactly on the fences and are typically not classified as outliers. The value 20 is between the fences (15 < 20 < 135), so it is not an outlier.
  7. If the range of a non-empty dataset is zero, which of the following statements must be true?
    Answer: The standard deviation of the dataset must be zero. The range of a dataset is the difference between the maximum and minimum values. If the range is zero, it means the maximum value is equal to the minimum value. This is only possible if all values in the dataset are identical. When all data points are the same, there is no variability or spread. Therefore, the standard deviation, which measures spread, must be zero. The mean does not have to be zero; for example, the dataset {5, 5, 5} has a range of 0 but a mean of 5. The IQR would also be zero, but it would not necessarily equal the mean.
  8. A dataset contains 10 distinct numbers. Which of the following operations, when performed on the dataset, must result in a new dataset with a standard deviation of 0?
    Answer: Multiplying each number by 0, then adding 10 to each result. A standard deviation of 0 means there is no spread in the data; all values in the dataset are identical. Let's analyze the options: A) Multiplying each number by 0 makes every value in the dataset 0. Then adding 10 makes every value 10. The resulting dataset is {10, 10, ..., 10}, which has a standard deviation of 0. B) The dataset of deviations (xi−xˉ)(x_i - \bar{x}) has a mean of 0, but its standard deviation is the same as the original dataset's standard deviation. C) Subtracting a constant (the median) from each number shifts the dataset but does not change its spread. The standard deviation remains unchanged. D) Percentile ranks will be different values (like 10th percentile, 20th percentile, etc.), so this creates a dataset with spread, not zero standard deviation.
  9. A dataset consists of four positive integers: {1, 2, 3, x}. The sample variance of this dataset is 503\frac{50}{3}. What is the value of x?
    Answer: 10 Let the mean be yˉ=1+2+3+x4=6+x4\bar{y} = \frac{1+2+3+x}{4} = \frac{6+x}{4}. The formula for sample variance is s2=1n−1∑(yi−yˉ)2s^2 = \frac{1}{n-1} \sum (y_i - \bar{y})^2. So, 503=13[(1−6+x4)2+(2−6+x4)2+(3−6+x4)2+(x−6+x4)2]\frac{50}{3} = \frac{1}{3} \left[ (1 - \frac{6+x}{4})^2 + (2 - \frac{6+x}{4})^2 + (3 - \frac{6+x}{4})^2 + (x - \frac{6+x}{4})^2 \right]. Multiplying both sides by 3 gives 50=∑(yi−yˉ)250 = \sum (y_i - \bar{y})^2. This algebraic path is complex. An alternative is to test the integer options. If x=10, the dataset is {1, 2, 3, 10}. The mean is 1+2+3+104=4\frac{1+2+3+10}{4} = 4. The deviations from the mean are (1-4)=-3, (2-4)=-2, (3-4)=-1, and (10-4)=6. The squared deviations are 9, 4, 1, and 36. The sum of squared deviations is 9+4+1+36=509+4+1+36=50. The sample variance is s2=504−1=503s^2 = \frac{50}{4-1} = \frac{50}{3}, which matches the given information.
  10. The heights of a sample of plants are measured in inches. The variance of these heights is found to be 9 square inches. If the heights are converted to feet by dividing each value by 12, what will be the new standard deviation of the heights, in feet?
    Answer: 0.25 This is a multi-step problem. First, find the original standard deviation from the variance. Standard deviation is the square root of variance, so the original standard deviation is 9=3\sqrt{9} = 3 inches. Second, apply the unit conversion. Dividing all data points by a constant (12) also divides their standard deviation by that constant. The new standard deviation in feet is 3 inches12=0.25\frac{3 \text{ inches}}{12} = 0.25 feet. Distractor A is the new variance (9/122=9/144=0.06259/12^2 = 9/144 = 0.0625), not the standard deviation. Distractor C results from incorrectly dividing the variance by 12 (9/12 = 0.75). Distractor D is the original standard deviation in inches.
  11. A small dataset is {3, 5, 7, 9, 11}. The standard deviation is calculated. Then, the maximum value, 11, is removed. What is the effect on the standard deviation?
    Answer: The standard deviation decreases because an extreme value was removed. When you encounter questions about how removing data points affects standard deviation, focus on understanding what standard deviation measures: the spread or variability of data around the mean. Let's examine the original dataset {3, 5, 7, 9, 11}. The mean is 7, and notice this data is perfectly symmetric around the mean. The standard deviation measures how far, on average, each value deviates from this mean. When you remove 11 (the maximum value), you're left with {3, 5, 7, 9}. The new mean becomes 6, and crucially, you've eliminated the data point that was farthest from the original center. This removal reduces the overall spread of the data, which decreases the standard deviation. Option A incorrectly suggests that smaller datasets automatically have larger standard deviations. Dataset size doesn't directly determine standard deviation magnitude – it's about the spread of values. Option B is wrong because this effect can be determined through logical reasoning about variability. Removing extreme values predictably reduces spread. Option C makes a flawed assumption about symmetry. While the original data was symmetric, removing the maximum value breaks this symmetry and, more importantly, reduces the range and variability regardless of the symmetry. Option D correctly identifies that removing an extreme value (one far from the mean) reduces the dataset's variability, thereby decreasing the standard deviation. Study tip: Remember that standard deviation reflects spread, not size. Removing outliers or extreme values typically decreases standard deviation, while removing values near the mean typically increases it.
  12. Two datasets, Set A and Set B, are compared. Every value in Set B is exactly 10 units greater than the corresponding value in Set A. Which of the following statements correctly describes the relationship between their measures of spread?
    Answer: The range and interquartile range of Set B are equal to those of Set A. Adding a constant value (10) to every data point is a linear transformation that shifts the entire distribution. Measures of center (like the mean and median) will increase by 10. However, measures of spread (range, IQR, variance, standard deviation) are unaffected by addition or subtraction because the distances between the data points remain unchanged. Therefore, the range, IQR, variance, and standard deviation of Set B will be identical to those of Set A.
  13. A researcher calculates the variance for a sample of 10 items and gets a value of s2=20s^2 = 20. She later realizes this sample represents the entire population of interest. What is the true population variance, σ2\sigma^2?
    Answer: 18.0 The formula for sample variance is s2=∑(xi−xˉ)2n−1s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1}, while the formula for population variance is σ2=∑(xi−xˉ)2n\sigma^2 = \frac{\sum(x_i - \bar{x})^2}{n}. The key is the sum of squared deviations, ∑(xi−xˉ)2\sum(x_i - \bar{x})^2. From the sample variance, we can find this sum: ∑(xi−xˉ)2=s2(n−1)=20(10−1)=20(9)=180\sum(x_i - \bar{x})^2 = s^2(n-1) = 20(10-1) = 20(9) = 180. Now, we can calculate the population variance using this sum and dividing by nn: σ2=18010=18.0\sigma^2 = \frac{180}{10} = 18.0.
  14. For a given dataset, the variance is 64. What is the correct interpretation of the standard deviation?
    Answer: The standard deviation is 8, representing a typical deviation from the mean. When you encounter questions about variance and standard deviation, remember that these two measures of spread are mathematically related but have different units and interpretations. Since variance is 64, the standard deviation is 64=8\sqrt{64} = 8. Standard deviation is always the square root of variance. More importantly, standard deviation represents how much individual data points typically deviate from the mean, measured in the same units as the original data. This makes it much more interpretable than variance. Let's examine why each option is correct or incorrect: Option A is wrong because it correctly calculates the standard deviation as 8, but incorrectly describes it as "squared spread." Variance (not standard deviation) involves squared units. Option B is incorrect because it claims standard deviation equals variance. These are never the same value unless variance equals 0 or 1, which are special cases. Option C is wrong because it suggests standard deviation is 642=409664^2 = 4096. This reverses the relationship—standard deviation is the square root of variance, not variance squared. Option D is correct because it properly calculates standard deviation as 8 and accurately describes its meaning: a typical deviation from the mean. Study tip: Remember that standard deviation = variance\sqrt{\text{variance}}, and standard deviation is always more interpretable because it's in the original units of measurement. When you see variance, immediately think about taking its square root to get the standard deviation for easier interpretation of spread.
  15. The mean daily high temperature in a city during summer is 85°F with a standard deviation of 5°F. The shape of the temperature distribution is unknown. According to Chebyshev's inequality, at least what percentage of days must have a temperature between 70°F and 100°F?
    Answer: 89% Chebyshev's inequality can be used for any distribution. First, determine how many standard deviations from the mean the interval extends. The interval is from 85−1585 - 15 to 85+1585 + 15. The distance from the mean is 15°F. With a standard deviation of 5°F, this distance represents k=155=3k = \frac{15}{5} = 3 standard deviations. Chebyshev's inequality states that the proportion of data within kk standard deviations of the mean is at least 1−1k21 - \frac{1}{k^2}. For k=3k=3, this is 1−132=1−19=89≈0.8891 - \frac{1}{3^2} = 1 - \frac{1}{9} = \frac{8}{9} \approx 0.889, or 88.9%. The closest answer is 89%.
  16. A dataset consists of the following 5 numbers: {10, 12, 15, 18, 20}. A sixth number, 70, is added to the dataset. Which of the following measures of spread will experience the largest percentage increase due to the addition of this new number?
    Answer: Range The range is the most sensitive measure of spread to extreme outliers. Let's analyze the change: Original dataset: {10, 12, 15, 18, 20}. Range = 20 - 10 = 10. Q1=11, Q3=19, IQR=8. New dataset: {10, 12, 15, 18, 20, 70}. New Range = 70 - 10 = 60. The percentage increase is ((60-10)/10)*100% = 500%. New Q1=12, New Q3=20, New IQR=8. The percentage increase is 0%. While the standard deviation and variance will also increase, the range is directly defined by the extreme values and thus experiences the most dramatic percentage change when a significant outlier is introduced.
  17. The boxplot shown summarizes the scores of 200 students on a recent exam. What is the interquartile range of the scores?
    Answer: 20 The interquartile range (IQR) is the difference between the third quartile (Q3) and the first quartile (Q1). From the boxplot, Q3 is the right edge of the box, which is at 85. Q1 is the left edge of the box, which is at 65. Therefore, the IQR = Q3 - Q1 = 85 - 65 = 20. Distractor A (10) is the difference between Q3 and the median. Distractor C (45) is the difference between Q3 and the minimum. Distractor D (60) is the total range (Maximum - Minimum = 100 - 40).
  18. The two histograms shown summarize two datasets, A and B, each containing 500 observations. Both datasets have approximately the same mean and the same minimum and maximum values. Based on the histograms, which of the following statements is most likely true?
    Answer: The standard deviation of Dataset B is larger than that of Dataset A. Standard deviation measures the typical or average distance of data points from the mean. In Histogram A (bell-shaped), most data points are clustered near the mean, with fewer points in the tails. In Histogram B (uniform), the data points are spread out evenly across the entire range. This means that, on average, the points in Dataset B are farther from the mean than the points in Dataset A. Therefore, the standard deviation of Dataset B is larger. The interquartile range of the uniform distribution will be narrower than that of the bell-shaped distribution with the same range.

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