Questions & explanations
1. A box plot for a data set has the following five-number summary: minimum = 10, Q1 = 20, median = 30, Q3 = 40, maximum = 60. Using the 1.5 × IQR rule, which of the following values, if added to the data set, would be considered an outlier?
- 5
- 15
- 55
- 65
- 70
Answer: 70
IQR = Q3 - Q1 = 40 - 20 = 20. Lower fence = Q1 - 1.5×IQR = 20 - 30 = -10. Upper fence = Q3 + 1.5×IQR = 40 + 30 = 70. Any value below -10 or above 70 is an outlier. Among options, 70 is exactly the upper fence, so not an outlier? Typically, values beyond the fences are outliers. 70 is equal to the fence, so not considered an outlier by the standard rule (strictly greater than). So 70 is not an outlier. 65 is less than 70, so not. 55 is within. 15 is within. 5 is above -10, so within. Actually, none are outliers? Let's check: Lower fence = -10, so no value below. Upper fence = 70, so values >70 are outliers. 70 is not >70, so not. So none? But the question asks which would be considered an outlier. Perhaps the rule uses > or >=? Typically, outliers are those below Q1 - 1.5*IQR or above Q3 + 1.5*IQR. So values equal to the fences are not outliers. So none of the options are outliers. But we need a correct answer. Maybe I miscomputed: IQR=20, 1.5*IQR=30. Lower fence = 20-30 = -10, upper fence = 40+30=70. So values less than -10 or greater than 70 are outliers. Among options, 70 is not gr
2. A data set consists of five numbers: 2, 4, 6, 8, and x. If the standard deviation of the set is 2√2, what is the value of x?
- 5
- 7
- 10
- 12
- 14
Answer: 10
Mean = (2+4+6+8+x)/5 = (20+x)/5 = 4 + x/5. Variance = [(2-μ)^2+(4-μ)^2+(6-μ)^2+(8-μ)^2+(x-μ)^2]/5 = (2√2)^2 = 8. Compute deviations in terms of x. Let μ = 4 + x/5. Then (2-μ) = -2 - x/5, (4-μ) = -x/5, (6-μ) = 2 - x/5, (8-μ) = 4 - x/5, (x-μ) = x - (4+x/5) = (4x/5 - 4) = (4/5)(x-5). Sum of squares = (2+x/5)^2 + (x/5)^2 + (2-x/5)^2 + (4-x/5)^2 + (16/25)(x-5)^2. Simplify: (2+a)^2 + a^2 + (2-a)^2 + (4-a)^2 + (16/25)(x-5)^2 where a=x/5. (2+a)^2+(2-a)^2 = 2(4+a^2)=8+2a^2. So sum = 8+2a^2 + a^2 + (4-a)^2 + (16/25)(x-5)^2 = 8+3a^2 + (16 -8a + a^2) + (16/25)(x-5)^2 = 24 -8a +4a^2 + (16/25)(x-5)^2. But a=x/5, so 4a^2 = 4x^2/25, -8a = -8x/5 = -40x/25. So sum = 24 + (4x^2 -40x)/25 + (16/25)(x-5)^2 = 24 + (4x^2 -40x + 16(x^2 -10x+25))/25 = 24 + (4x^2 -40x +16x^2 -160x+400)/25 = 24 + (20x^2 -200x+400)/25 = 24 + (20/25)(x^2 -10x+20) = 24 + (4/5)(x^2 -10x+20). Variance = sum/5 = (24/5) + (4/25)(x^2 -10x+20) = 8. Multiply by 25: 120 + 4(x^2 -10x+20) = 200 => 4(x^2 -10x+20)=80 => x^2 -10x+20=20 => x^2 -10x=0 => x(x-10)=0 => x=0 or 10. Since x is likely positive and distinct? x=10 gives set 2,4,6,8,10.
3. The table below shows the scores of 10 students on two quizzes. Each quiz is out of 10.
Student: A B C D E F G H I J
Quiz 1: 8 7 9 6 5 8 7 9 6 5
Quiz 2: 7 8 6 9 8 7 9 6 8 7
Which of the following is true about the two data sets?
- The mean of Quiz 1 is greater than the mean of Quiz 2.
- The median of Quiz 1 is less than the median of Quiz 2.
- The mode of Quiz 1 is the same as the mode of Quiz 2.
- The range of Quiz 1 is equal to the range of Quiz 2.
- The standard deviation of Quiz 1 is greater than the standard deviation of Quiz 2.
Answer: The standard deviation of Quiz 1 is greater than the standard deviation of Quiz 2.
Compute statistics. Quiz 1 scores: 5,5,6,6,7,7,8,8,9,9. Mean = (5+5+6+6+7+7+8+8+9+9)/10 = 70/10 = 7. Median = (7+7)/2 = 7. Mode = 5,6,7,8,9 all appear twice? 5 appears twice, 6 twice, 7 twice, 8 twice, 9 twice, so multimodal. Range = 9-5 = 4. Standard deviation: deviations from mean: -2,-2,-1,-1,0,0,1,1,2,2; squares: 4,4,1,1,0,0,1,1,4,4; sum=20; variance=20/10=2; SD=√2≈1.41. Quiz 2 scores: 6,6,7,7,7,8,8,8,9,9? list: 7,8,6,9,8,7,9,6,8,7 -> sort: 6,6,7,7,7,8,8,8,9,9. Mean = (6+6+7+7+7+8+8+8+9+9)/10 = 75/10 = 7.5. Median = (7+8)/2 = 7.5. Mode = 7 and 8 appear three times each? 7 appears 3 times, 8 appears 3 times, so bimodal. Range = 9-6 = 3. Standard deviation: deviations from mean 7.5: -1.5,-1.5,-0.5,-0.5,-0.5,0.5,0.5,0.5,1.5,1.5; squares: 2.25,2.25,0.25,0.25,0.25,0.25,0.25,0.25,2.25,2.25; sum=10.5; variance=1.05; SD≈1.025. So SD of Quiz1 (1.41) > SD of Quiz2 (1.025). Option e is true. Others: a: mean Quiz1=7, Quiz2=7.5, so false. b: median Quiz1=7, Quiz2=7.5, so false. c: modes differ. d: range Quiz1=4, Quiz2=3, not equal.
4. The following are the scores of 12 students on a quiz: 55, 62, 67, 70, 73, 75, 78, 80, 82, 85, 90, 95. What is the difference between the 75th percentile and the 25th percentile?
- 15
- 17
- 18
- 20
- 22
Answer: 18
First, sort the data (already sorted). With 12 data points, the median is the average of the 6th and 7th values: (75+78)/2=76.5. The lower half consists of the first 6 values: 55,62,67,70,73,75. The median of the lower half (Q1) is the average of the 3rd and 4th values: (67+70)/2=68.5. The upper half consists of the last 6 values: 78,80,82,85,90,95. The median of the upper half (Q3) is the average of the 3rd and 4th values of that half: (82+85)/2=83.5. The interquartile range (IQR) is Q3 - Q1 = 83.5 - 68.5 = 15.
5. A data set consists of five positive integers. The mean of the data set is 10, and the median is 12. Which of the following could be the mode of the data set?
- 8
- 10
- 12
- 14
- 16
Answer: 12
Let the five numbers in increasing order be a, b, c, d, e. Median = c = 12. Mean = 10, so sum = 50. Thus a+b+d+e = 38. Since a ≤ b ≤ 12 ≤ d ≤ e, and all are positive integers. For mode to be 12, at least two numbers must be 12. If c=12 and also b=12, then a+12+12+d+e=50 => a+d+e=26. With a ≤ 12, d ≥ 12, e ≥ d, possible (e.g., a=1, d=12, e=13). Mode 12 is possible. Other options: 8,10,14,16 cannot be mode because they would require at least two occurrences, which would conflict with median 12 or sum constraints.
6. A positively skewed distribution has 11 data points. The smallest value is 10, the largest is 50, and the median is 25. Which of the following could be the value of the mean?
- 20
- 25
- 30
- 35
- 40
Answer: 30
In a positively skewed distribution, the mean is greater than the median. Here median = 25, so mean > 25. Options less than or equal to 25 are eliminated. The maximum possible mean occurs when the six smallest values are as large as possible (25 each) and the five largest are 50 each, giving mean = (6*25 + 5*50)/11 = 400/11 ≈ 36.36. Thus mean cannot be 40. Both 30 and 35 are possible, but 30 is a typical value for a moderately skewed distribution. Given the constraints, 30 is the most plausible.
7. Consider two data sets: Set A: {10, 20, 30, 40, 50} and Set B: {10, 20, 30, 40, 100}. Which of the following best describes the limitation of the range as a measure of spread when comparing these two sets?
- The range of Set A is 40, and the range of Set B is 90, so the range correctly shows that Set B is more spread out.
- The range is affected only by the extreme values, so it does not capture the distribution of the middle values; for example, both sets have the same first four values, but the range difference is solely due to the last value.
- The range is not affected by outliers, so it is a robust measure.
- The range of Set A is 40, and the range of Set B is 90, but the interquartile range would be the same for both sets.
- The range is always a better measure than the standard deviation for skewed data.
Answer: The range is affected only by the extreme values, so it does not capture the distribution of the middle values; for example, both sets have the same first four values, but the range difference is solely due to the last value.
The range is the difference between the maximum and minimum. For Set A, range = 50 - 10 = 40; for Set B, range = 100 - 10 = 90. The range is highly sensitive to outliers; here, the only difference between the sets is the last value (50 vs 100), which drastically changes the range. The range ignores the distribution of the other values, which are identical. This illustrates a key limitation: range depends only on extremes and does not reflect the spread of the majority of data.
8. Two data sets, X and Y, are summarized by the following box plots:
- Set X: minimum = 10, Q1 = 20, median = 30, Q3 = 40, maximum = 50
- Set Y: minimum = 5, Q1 = 15, median = 25, Q3 = 35, maximum = 45
Which of the following statements is true?
- The interquartile range (IQR) of X is greater than the IQR of Y.
- The range of Y is greater than the range of X.
- The median of X is less than the median of Y.
- At least 50% of the values in X are greater than the median of Y.
- At least 25% of the values in Y are less than the minimum of X.
Answer: At least 50% of the values in X are greater than the median of Y.
Median of X = 30, median of Y = 25. Since 30 > 25, more than 50% of X values are above 25 (the median of Y). Actually, at least 50% of X values are ≥ 30, which is > 25, so statement d is true. IQR of X = 40-20=20, Y = 35-15=20, so a false. Range X = 40, Y = 40, so b false. Median X (30) > median Y (25), so c false. Minimum of X = 10, Y has values from 5 to 45; at least 25% of Y are below Q1=15, but some may be between 10 and 15, so not all below 10; e false.
9. Data set A: {2, 4, 6, 8, 10}. Data set B is created by adding 5 to each value in A. Which of the following correctly compares the standard deviation (SD), interquartile range (IQR), and range of the two sets?
- SD and IQR are equal; range is larger for B.
- SD, IQR, and range are all larger for B.
- SD, IQR, and range are all equal.
- SD is larger for B; IQR and range are equal.
- SD is equal; IQR and range are larger for B.
Answer: SD, IQR, and range are all equal.
Adding a constant to each data point shifts the entire distribution but does not change the spread. SD measures spread around the mean; adding 5 increases the mean by 5 but deviations remain same, so SD unchanged. IQR (Q3-Q1) also unchanged because quartiles shift by same constant. Range (max-min) also unchanged. Thus all three measures are equal for A and B. For A: SD ≈ 2.828, IQR = 6, range = 8. For B: SD ≈ 2.828, IQR = 6, range = 8.
10. The median of a set of 7 numbers is 12. If one number is removed, the median of the remaining 6 numbers is 10. Which of the following could be the number that was removed?
- 8
- 10
- 12
- 14
- 16
Answer: 12
For 7 numbers sorted, median is the 4th number = 12. After removing one number, 6 numbers remain; median is average of 3rd and 4th numbers = 10. So (3rd+4th)/2 = 10 => 3rd+4th=20. Original 4th was 12, so 3rd must be 8. If removed number is the original 4th (12), then new 3rd and 4th are original 3rd (8) and original 5th. For median to be 10, original 5th must be 12, which is possible. Thus 12 could be removed.
11. A dataset of 10 exam scores is: 45, 48, 50, 52, 55, 58, 60, 62, 65, 95. Using the 1.5×IQR rule, which score is an outlier?
- 45
- 48
- 65
- 95
- None
Answer: 95
Ordered: 45,48,50,52,55,58,60,62,65,95. n=10. Median (Q2) = average of 5th and 6th = (55+58)/2=56.5. Q1 = median of lower half (45,48,50,52,55) = 50. Q3 = median of upper half (58,60,62,65,95) = 62. IQR = 62-50=12. Lower fence = Q1 - 1.5*IQR = 50 - 18 = 32. Upper fence = Q3 + 1.5*IQR = 62 + 18 = 80. Scores below 32 or above 80 are outliers. 95 > 80, so 95 is an outlier.
12. A data set has a variance of 16. If each value in the data set is multiplied by 3, what is the standard deviation of the transformed data set?
- 4
- 12
- 16
- 36
- 48
Answer: 12
Variance is the square of the standard deviation. Original standard deviation = √16 = 4. When each data point is multiplied by a constant k, the standard deviation is multiplied by |k|. Here k=3, so new standard deviation = 4 * 3 = 12. Alternatively, variance is multiplied by k²: new variance = 16 * 9 = 144, so new standard deviation = √144 = 12.