CAIIB ABM UNIT 3 MCQ – Measures of Central Tendency & Dispersion, Skewness, Kurtosis

Correct Answer: C. Classifying it into groups based on characteristics. Statistical data, once collected, is first organised by classifying it into different groups according to shared features and then presented in a table.

Correct Answer: C. Central Tendency and Dispersion. Central Tendency and Dispersion are widely used tools that handle large data quantities, simplifying them to a single value for comparative studies and clear conclusions.

Correct Answer: C. To condense a dataset into a single representative value. One of the main goals of measures of central tendency is to summarise or condense the data into one single value.

Correct Answer: C. Central Tendency. Central tendency refers to the characteristic of data points grouping or clustering around a central or midpoint value.

Correct Answer: C. Rigidly. A good measure of central tendency must be rigidly defined, meaning it should be clear, unambiguous, and have a precise interpretation.

Correct Answer: B. It should be simple to understand and easy to calculate. An ideal measure of central tendency should be straightforward for users to comprehend and compute without excessive difficulty.

Correct Answer: C. It should be capable of further mathematical treatment. A good measure should allow for its use in subsequent mathematical or algebraic operations and analyses.

Correct Answer: C. It should not be unduly affected by the extreme values. An effective measure of central tendency should remain relatively stable and representative, even in the presence of unusually high or low values in the data.

Correct Answer: B. Mean, Median, Mode. Mean, Median, and Mode as the three principal types of averages used to measure central tendency.

Correct Answer: D. Mean. The mean (or average) is highlighted as the most frequently employed single measure, known for its simplicity in calculation, understanding, and interpretation. Mean is of three types: Arithmetic Mean, Geometric Mean and Harmonic Mean.

Correct Answer: C. By summing all the numbers and dividing by the count of the numbers. The arithmetic mean (AM) for ungrouped data is found by adding all the individual values together and then dividing this total sum by the number of values in the set. For example, for the numbers 2, 4, 6, the sum is 2 + 4 + 6 = 12. There are 3 numbers. So, the AM = 12 / 3 = 4. The formula is AM = (Σxᵢ) / n, where Σxᵢ is the sum of all observations and n is the number of observations.

Correct Answer: C. The class mark or midpoint of each class interval. For grouped data, especially with class intervals, xᵢ represents the midpoint (or class mark) of the i-th class interval. This midpoint is used as the representative value for all observations falling within that class. For example, if a class interval is 20-30, the class mark xᵢ would be (20 + 30) / 2 = 25. This value is then multiplied by the frequency (fᵢ) of that class.

Correct Answer: B. The arithmetic mean and size (number of observations) of each individual group. To find the combined mean of multiple groups, you need to know the arithmetic mean (e.g., X̄₁, X̄₂) and the number of observations (size, e.g., n₁, n₂) for each group. The formula involves weighting each group’s mean by its size: Combined AM = (n₁X̄₁ + n₂X̄₂) / (n₁ + n₂). For instance, if Group A has 10 students (n₁) with an average score of 70 (X̄₁) and Group B has 20 students (n₂) with an average score of 80 (X̄₂), the combined mean is (1070 + 2080) / (10 + 20) = (700 + 1600) / 30 = 2300 / 30 = 76.67.

Correct Answer: B. By subtracting the incorrect value and adding the correct value to the original sum, then dividing by the number of observations. First, find the original (incorrect) sum of observations using Sum = Incorrect Mean * n. Then, adjust this sum by subtracting the wrongly recorded value and adding the actual, correct value. Finally, divide this corrected sum by the number of observations (n) to get the corrected mean. For example, if the mean of 5 observations was 10 (Sum = 50) and 8 was misread as 3, the corrected sum = 50 – 3 + 8 = 55. The corrected mean = 55 / 5 = 11.

Correct Answer: C. It is based on all the observations in the dataset. A key advantage of the arithmetic mean is that its calculation incorporates every single value in the dataset, making it representative of the entire data distribution.

Correct Answer: D. It is highly affected by extreme values. A major drawback of the arithmetic mean is its sensitivity to extreme values (outliers). A single very high or very low value can pull the mean significantly towards it, potentially making it less representative of the central tendency of the bulk of the data. For example, the mean of 2, 3, 4, 5, 100 is (2+3+4+5+100)/5 = 114/5 = 22.8, which is much higher than most numbers in the set due to the extreme value 100.

Correct Answer: A. 20. The sum of marks is 15 + 25 + 30 + 10 + 20 = 100. The number of students (observations) is 5. Arithmetic Mean = Sum of observations / Number of observations = 100 / 5 = 20.

Correct Answer: C. 4.18. We calculate Σfᵢxᵢ = (32) + (54) + (26) + (18) = 6 + 20 + 12 + 8 = 46. The total frequency Σfᵢ = 3 + 5 + 2 + 1 = 11. Arithmetic Mean = Σfᵢxᵢ / Σfᵢ = 46 / 11 ≈ 4.18.

Correct Answer: A. 25. First, find the class marks (midpoints): (10+20)/2 = 15; (20+30)/2 = 25; (30+40)/2 = 35. Calculate Σfᵢxᵢ = (515) + (1025) + (5*35) = 75 + 250 + 175 = 500. Total frequency Σfᵢ = 5 + 10 + 5 = 20. Arithmetic Mean = Σfᵢxᵢ / Σfᵢ = 500 / 20 = 25.

Correct Answer: B. 66. Here, n₁=40, X̄₁=60, n₂=60, X̄₂=70. Combined Mean = (n₁X̄₁ + n₂X̄₂) / (n₁ + n₂) = (4060 + 6070) / (40 + 60) = (2400 + 4200) / 100 = 6600 / 100 = 66.

Correct Answer: A. 23. Given Mean (X̄) = 30 and number of students (n) = 10. The sum of observations Σxᵢ = n * X̄ = 10 * 30 = 300. The sum of the known marks is 35+30+18+15+40+30+52+47+10 = 277. So, 277 + x = 300. Therefore, x = 300 – 277 = 23.

Correct Answer: C. Rs. 689. Original Mean = 680, n = 50. Original Sum = Mean * n = 680 * 50 = 34000. Incorrect value = 270, Correct value = 720. Corrected Sum = Original Sum – Incorrect Value + Correct Value = 34000 – 270 + 720 = 34450. Corrected Mean = Corrected Sum / n = 34450 / 50 = 689.

Correct Answer: B. The nth root of the product of the n numbers. The Geometric Mean (GM) measures central tendency by considering the compounding effect. It is calculated by multiplying all the n positive numbers together and then taking the nth root of the product. For example, the GM of 2 and 8 is the square root of (2 ⋅ 8), which is √16 = 4. The formula is GM = (x₁ ⋅ x₂ ⋅ … ⋅ x<0xE2><0x82><0x99>)¹⁄ₙ or ⁿ√(x₁ ⋅ x₂ ⋅ … ⋅ x<0xE2><0x82><0x99>).

Correct Answer: C. Averaging rates of change, like compound annual growth rates or stock index returns. GM is suitable for averaging ratios or rates of change over time because it accounts for compounding. For example, if an investment grows by 10% in year 1 (factor 1.10) and 20% in year 2 (factor 1.20), the average annual growth factor is the GM of 1.10 and 1.20, which is √(1.10 ⋅ 1.20) = √1.32 ≈ 1.149, representing an average growth rate of 14.9%.

Correct Answer: C. It cannot be calculated if any value in the dataset is zero or negative. The calculation of GM involves taking the product of all values. If any value is zero, the product becomes zero. If any value is negative, taking an even root (like square root) of the product might result in an imaginary number or be undefined in the context of real-valued means. Thus, GM is restricted to positive values only.

Correct Answer: C. The reciprocal of the arithmetic mean of the reciprocals of the observations. To find the HM, you first take the reciprocal of each observation, then calculate the arithmetic mean of these reciprocals, and finally, take the reciprocal of that result. For example, the HM of 2 and 4 is: Reciprocals are ¹⁄₂ and ¹⁄₄. Their AM is (¹⁄₂ + ¹⁄₄) / 2 = (³⁄₄) / 2 = ³⁄₈. The HM is the reciprocal of ³⁄₈, which is ⁸⁄₃ ≈ 2.67. The formula is HM = n / ∑(¹⁄xᵢ).

Correct Answer: C. When averaging ratios or rates, such as speeds over equal distances. HM is suitable when averaging rates, which are ratios of two different units (like km/hr). For instance, if a car travels a certain distance at 40 km/hr and returns the same distance at 60 km/hr, the average speed is the HM of 40 and 60. HM = 2 / (¹⁄₄₀ + ¹⁄₆₀) = 2 / ((3+2)⁄120) = 2 / (⁵⁄₁₂₀) = 2 ⋅ (¹²⁰⁄₅) = ²⁴⁰⁄₅ = 48 km/hr.

Correct Answer: D. AM ⋅ HM = GM². For any set of positive observations, the Arithmetic Mean, Geometric Mean, and Harmonic Mean are related by the equation: Arithmetic Mean multiplied by Harmonic Mean equals the square of the Geometric Mean. AM > HM > GM.

Correct Answer: A. 18.42. The GM is the 4th root of the product of the numbers: GM = (10 ⋅ 24 ⋅ 15 ⋅ 32)¹⁄₄ = (115200)¹⁄₄ ≈ 18.423.

Correct Answer: C. 16.69. The HM is calculated as n / ∑(¹⁄xᵢ). Here n=4. ∑(¹⁄xᵢ) = ¹⁄₁₀ + ¹⁄₂₄ + ¹⁄₁₅ + ¹⁄₃₂ = 0.1000 + 0.0417 + 0.0667 + 0.0313 = 0.2397 (approx). HM = 4 / 0.2397 ≈ 16.69.

Correct Answer: B. 2.47. The formula for grouped data is GM = (x₁^f₁ ⋅ x₂^f₂ ⋅ … ⋅ )¹⁄ᴺ, where N = ∑f. N = 5+6+5+10 = 26. GM = (1⁵ ⋅ 2⁶ ⋅ 3⁵ ⋅ 4¹⁰)¹⁄₂₆ = (1 ⋅ 64 ⋅ 243 ⋅ 1048576)¹⁄₂₆ = (16220606464)¹⁄₂₆. Using logarithms or a calculator gives GM ≈ 2.4705.

Correct Answer: B. 4. Using the relationship AM ⋅ HM = GM². We are given AM = 16 and GM = 8. So, 16 ⋅ HM = 8². 16 ⋅ HM = 64. HM = 64 / 16 = 4.

Correct Answer: C. The value that divides the distribution into two equal parts. The median is the middle value in a dataset that has been arranged in order of magnitude. It splits the data such that half the observations are below it and half are above it. For example, in the ordered set {2, 3, 7, 8, 11}, the median is 7, as it’s the middle value with two observations below and two above.

Correct Answer: C. Arrange the observations in ascending or descending order. The median is a positional average, so finding its value requires the data to be sorted first, either from smallest to largest (ascending) or largest to smallest (descending). Without ordering, the identified ‘middle’ value would be arbitrary.

Correct Answer: C. Mean is based on all observations, while the median depends on position. The arithmetic mean is calculated using the value of every observation in the dataset. In contrast, the median is determined by its position within the ordered dataset (the middle value), making it a positional average rather than a calculated one.

Correct Answer: B. To divide the data into four equal parts. Quartiles are points that divide the ordered data distribution into four segments, each containing approximately 25% of the observations. These points are the first quartile (Q1), the second quartile (Q2, which is the median), and the third quartile (Q3).

Correct Answer: B. The difference between the third quartile (Q3) and the first quartile (Q1). The Interquartile Range is a measure of statistical dispersion, specifically representing the range within which the central 50% of the data lies. It’s calculated as IQR = Q3 − Q1. For example, if Q1 is 20 and Q3 is 50, the IQR is 50 − 20 = 30.

Correct Answer: B. The value of the ((n+1)⁄2)th observation in the ordered list. When there’s an odd number of observations after arranging them, the median is simply the value of the observation located exactly in the middle. For example, if n=9, the median is the (9+1)⁄2 = 5th observation in the sorted list.

Correct Answer: B. The average of the (n⁄2)th and ((n⁄2) + 1)th observations in the ordered list. When there’s an even number of observations, there isn’t a single middle value. The median is calculated by taking the arithmetic mean of the two central observations. For example, if n=10, the median is the average of the (10⁄2)=5th and ((10⁄2)+1)=6th observations in the sorted list.

Correct Answer: C. It is not significantly affected by extreme values (outliers). Since the median’s value depends only on the value(s) in the middle of the ordered data, it is robust to extreme high or low values at the ends of the distribution. For example, the median of {2, 3, 7, 8, 100} is 7, which is not pulled upwards by the extreme value 100, unlike the mean.

Correct Answer: C. It is not based on all the observations in the dataset. The median calculation only uses the middle one or two values of the ordered dataset, ignoring the specific magnitudes of the other observations. This means it doesn’t utilize all the information available in the data, unlike the mean.

Correct Answer: C. Because it is not capable of further algebraic treatment. Unlike the mean, the median cannot be easily used in further mathematical formulas or algebraic manipulations (e.g., calculating a combined median from the medians of subgroups is complex and often not exact).

Correct Answer: B. 32. First, arrange the data in ascending order: 21, 24, 27, 30, 32, 34, 35, 38, 48. There are n=9 observations (odd). The median is the ((n+1)⁄2)th observation, i.e., the (9+1)⁄2 = 5th observation. The 5th observation is 32.

Correct Answer: C. 16. Arrange the wages in ascending order: 8, 9, 11, 14, 15, 17, 18, 20, 22, 25. There are n=10 observations (even). The median is the average of the (n⁄2)th and ((n⁄2)+1)th observations, i.e., the average of the 5th and 6th observations. The 5th observation is 15, and the 6th is 17. Median = (15 + 17) / 2 = 16.

Correct Answer: B. 40-50. Total frequency N = 8+26+30+20+16 = 100. Calculate N⁄2 = 100⁄2 = 50. Find the cumulative frequencies (cf): 8, (8+26)=34, (34+30)=64, (64+20)=84, (84+16)=100. The median class is the class whose cumulative frequency is the first to just exceed N⁄2 = 50. The cf 64 is the first to exceed 50, which corresponds to the class interval 40-50.

Correct Answer: C. 45.33. From Q45, N=100, N⁄2=50, Median Class is 40-50. Lower limit l₁ = 40. Upper limit l₂ = 50. Frequency of median class f = 30. Cumulative frequency of preceding class cf = 34. Median = l₁ + {(l₂ − l₁) ⋅ (N⁄2 − cf)} / f = 40 + {(50 − 40) ⋅ (50 − 34)} / 30 = 40 + {10 ⋅ 16} / 30 = 40 + 160 / 30 = 40 + 5.33 = 45.33.

Correct Answer: B. 6. Calculate cumulative frequencies (cf): 14, 34, 74, 128, 168, 186, 193, 200. Total frequency N = 200. Calculate N⁄2 = 200⁄2 = 100. Find the cf just exceeding 100, which is 128. The value of X corresponding to cf=128 is 6. Therefore, the median is 6.

Correct Answer: A. 18.85. Adjust intervals to be continuous: 4.5-9.5, 9.5-14.5, 14.5-19.5, 19.5-24.5, 24.5-29.5, 29.5-34.5, 34.5-39.5. Calculate cf: 8, 26, 53, 74, 84, 92, 99. N = 99. N⁄2 = 99⁄2 = 49.5. The cf just exceeding 49.5 is 53, corresponding to the median class 14.5-19.5. l₁ = 14.5, l₂ = 19.5, f = 27, cf (preceding) = 26. Median = 14.5 + {(19.5 − 14.5) ⋅ (49.5 − 26)} / 27 = 14.5 + {5 ⋅ 23.5} / 27 = 14.5 + 117.5 / 27 = 14.5 + 4.35 = 18.85.

Correct Answer: D. 25. Since the median is 24, the median class is 20-30. Total frequency N = 5+25+a+18+7 = 55+a. l₁=20, l₂=30, f=a, cf (preceding) = 5+25 = 30. Median = l₁ + {(l₂ − l₁) ⋅ (N⁄2 − cf)} / f. 24 = 20 + {(30 − 20) ⋅ ((55+a)⁄2 − 30)} / a. 4 = {10 ⋅ ((55+a − 60)⁄2)} / a. 4a = 10 ⋅ (a−5) / 2. 4a = 5 ⋅ (a−5). 4a = 5a − 25. a = 25.

Correct Answer: B. 21.39. N = 200. For Q1, we need N⁄4 = 200⁄4 = 50. Calculate cf: 12, 40, 76, 126, 151, 169, 185, 195, 200. The cf just exceeding 50 is 76, which corresponds to the class 20-25. Q1 Class is 20-25. l₁=20, l₂=25, f=36, cf (preceding) = 40. Q1 = l₁ + {(l₂ − l₁) ⋅ (N⁄4 − cf)} / f = 20 + {(25 − 20) ⋅ (50 − 40)} / 36 = 20 + {5 ⋅ 10} / 36 = 20 + 50 / 36 = 20 + 1.39 = 21.39.

Correct Answer: A. 34.8. N = 200. For Q3, we need 3N⁄4 = 3⋅200⁄4 = 150. Calculate cf: 12, 40, 76, 126, 151, 169, 185, 195, 200. The cf just exceeding 150 is 151, which corresponds to the class 30-35. Q3 Class is 30-35. l₁=30, l₂=35, f=25, cf (preceding) = 126. Q3 = l₁ + {(l₂ − l₁) ⋅ (3N⁄4 − cf)} / f = 30 + {(35 − 30) ⋅ (150 − 126)} / 25 = 30 + {5 ⋅ 24} / 25 = 30 + 120 / 25 = 30 + 4.8 = 34.8.

Correct Answer: A. 13.41. The Interquartile Range is defined as IQR = Q3 − Q1. Using the calculated values, IQR = 34.8 − 21.39 = 13.41.

Correct Answer: C. The value that occurs most frequently. The mode is the value or number within a dataset that appears more times than any other value. For example, in the dataset {2, 3, 5, 3, 4, 3, 6}, the mode is 3 because it occurs three times, which is more than any other number.

Correct Answer: C. A distribution where two or more values occur with the highest frequency. If a dataset has two values that share the highest frequency of occurrence, it is called bimodal. If it has more than two such values, it is called multimodal. For instance, in {2, 2, 3, 4, 4, 5}, both 2 and 4 are modes, making it bimodal.

Correct Answer: C. It is easy to calculate and understand. One of the main advantages of the mode is its simplicity; it is often easy to identify by inspection (for ungrouped data) and straightforward to comprehend as the ‘most typical’ value.

Correct Answer: C. It is ill-defined as it is not based on all observations. A major criticism of the mode is that it only considers the most frequent value(s) and ignores the rest of the data, making it potentially unrepresentative of the overall dataset and thus considered ill-defined. It may not even exist or be unique.

Correct Answer: C. Mode = 3 Median – 2 Mean. For distributions that are not symmetrical but not extremely skewed, there is an approximate relationship often observed: Mode ≈ 3 * Median – 2 * Mean. This formula can be used to estimate one measure if the other two are known.

Correct Answer: D. 23. By observing the data, the value 23 appears 3 times. The value 27 appears 2 times. All other values appear only once. Since 23 occurs most frequently, the mode is 23.

Correct Answer: C. 40-50. The modal class is the class interval with the highest frequency. In this distribution, the highest frequency is 30, which corresponds to the class interval 40-50.

Correct Answer: A. 42.86. Modal class is 40-50. Lower limit l₁ = 40. Upper limit l₂ = 50. Frequency of modal class f₁ = 30. Frequency of preceding class f₀ = 26. Frequency of succeeding class f₂ = 20. Formula: Mode = l₁ + {(l₂ – l₁) * (f₁ – f₀)} / {2f₁ – f₀ – f₂} = 40 + {(50 – 40) * (30 – 26)} / {2*30 – 26 – 20} = 40 + {10 * 4} / {60 – 46} = 40 + 40 / 14 = 40 + 2.86 = 42.86.

Correct Answer: B. 80. Since the mode is 115, the modal class is 105-120. l₁ = 105, l₂ = 120, f₁ = a, f₀ = 60, f₂ = 70. Mode = l₁ + {(l₂ – l₁) * (f₁ – f₀)} / {2f₁ – f₀ – f₂}. 115 = 105 + {(120 – 105) * (a – 60)} / {2a – 60 – 70}. 10 = {15 * (a – 60)} / {2a – 130}. 10 * (2a – 130) = 15 * (a – 60). 20a – 1300 = 15a – 900. 5a = 400. a = 80.

Correct Answer: A. 23. Using the empirical relationship Mode = 3 Median – 2 Mean. Given Median = 25 and Mean = 26. Mode = (3 * 25) – (2 * 26) = 75 – 52 = 23.

Correct Answer: B. The degree of variability or spread of individual items. Measures of dispersion quantify how much the individual data points in a set are scattered, stretched, or spread out from the average or central value. For example, the sets {10, 20, 30} and {19, 20, 21} both have a mean of 20, but the first set has greater dispersion (spread).

Correct Answer: C. Absolute measures express dispersion in the original units of data, while relative measures are unitless ratios used for comparison. Absolute measures (like Range, QD, MD, SD) provide the dispersion in the same units as the data (e.g., rupees, kg). Relative measures (like Coefficient of Range, Coefficient of QD, etc.) are ratios or percentages, making them unit-free and suitable for comparing the variability of datasets with different units or different average values.

Correct Answer: B. By finding the difference between the highest value (Maximum) and the lowest value (Minimum). The range is simply the difference between the largest and smallest observation in the dataset. For example, in the dataset {15, 18, 20, 22}, the Maximum is 22 and the Minimum is 15, so the Range = 22 – 15 = 7.

Correct Answer: C. It is heavily influenced by extreme values and ignores the spread of intermediate values. The range only considers the two extreme points (maximum and minimum) and disregards how the rest of the data is distributed. A single outlier can drastically change the range, making it potentially unrepresentative of the overall variability.

Correct Answer: B. Half the difference between the third quartile (Q3) and the first quartile (Q1). Quartile Deviation measures the average range of the middle 50% of the data. It is calculated as QD = (Q3 – Q1) / 2. For example, if Q1 = 20 and Q3 = 50, then QD = (50 – 20) / 2 = 15.

Correct Answer: C. It is not affected by extreme values. Since QD is calculated using Q1 and Q3, it focuses on the spread of the central 50% of the data and is therefore not influenced by the extreme minimum or maximum values, unlike the Range.

Correct Answer: B. The average of the absolute deviations from a measure of central tendency (like mean, median, or mode). Mean Deviation calculates the average amount by which observations deviate from a central point (commonly the mean or median), ignoring the sign (positive or negative) of the deviation. For example, deviations from mean for {2, 3, 7} (mean=4) are |-2|, |-1|, |3|. MD = (2+1+3)/3 = 2.

Correct Answer: B. To ensure the sum of deviations is not always zero. The sum of simple deviations (without ignoring the sign) from the arithmetic mean is always zero (Σ(xᵢ – x̄) = 0). Taking the absolute value (ignoring the minus signs) ensures that all deviations contribute positively to the measure of dispersion.

Correct Answer: B. Median. A key property of Mean Deviation is that the sum of the absolute deviations (and thus the MD itself) is minimized when the deviations are taken from the median of the distribution.

Correct Answer: C. Ignoring the algebraic signs (+/-) of deviations makes it unsuitable for further algebraic treatment. The process of taking absolute values is mathematically inconvenient and limits the use of Mean Deviation in more advanced statistical analysis compared to measures like Standard Deviation.

Correct Answer: B. To obtain a unitless measure for comparing the variability of different datasets. Coefficients of dispersion are relative measures, typically calculated by dividing the absolute measure (e.g., Range, QD, MD) by an appropriate average (e.g., Mean, Median, Mid-range). This creates a unit-free ratio or percentage, allowing for comparison of spread between datasets with different units or averages.

Correct Answer: C. Studying economic issues like income distribution. The text specifically mentions that Mean Deviation and its coefficient are used in studying economic problems such as the distribution of income and wealth within a society.

Correct Answer: C. 7. Range = Maximum Value – Minimum Value. Maximum = 22, Minimum = 15. Range = 22 – 15 = 7.

Correct Answer: A. 0.19. Coefficient of Range = (Maximum – Minimum) / (Maximum + Minimum). Max = 22, Min = 15. Coefficient = (22 – 15) / (22 + 15) = 7 / 37 ≈ 0.189 or 0.19.

Correct Answer: C. 15. Quartile Deviation (QD) = (Q3 – Q1) / 2. QD = (60 – 30) / 2 = 30 / 2 = 15.

Correct Answer: A. 0.33. Coefficient of QD = (Q3 – Q1) / (Q3 + Q1). Coefficient = (60 – 30) / (60 + 30) = 30 / 90 = 1/3 ≈ 0.33.

Correct Answer: A. 20.75. First, calculate the Mean (x̄) = (12+20+39+46+54+61+78+90) / 8 = 400 / 8 = 50. Then find absolute deviations from mean: |12-50|=38, |20-50|=30, |39-50|=11, |46-50|=4, |54-50|=4, |61-50|=11, |78-50|=28, |90-50|=40. Sum of absolute deviations = 38+30+11+4+4+11+28+40 = 166. MD(Mean) = Σ|xᵢ – x̄| / n = 166 / 8 = 20.75.

Correct Answer: A. 0.415. Coefficient of MD (Mean) = MD(Mean) / Mean. Coefficient = 20.75 / 50 = 0.415.

Correct Answer: A. 9.68. Class Marks (x): 25, 35, 45, 55, 65. Mean x̄ = 46. Calculate |xᵢ – x̄|: |25-46|=21, |35-46|=11, |45-46|=1, |55-46|=9, |65-46|=19. Calculate fᵢ|xᵢ – x̄|: (821)=168, (2611)=286, (301)=30, (209)=180, (16*19)=304. Sum Σfᵢ|xᵢ – x̄| = 168+286+30+180+304 = 968. Total Frequency N = 100. MD(Mean) = Σfᵢ|xᵢ – x̄| / N = 968 / 100 = 9.68.

Correct Answer: C. 0.2104. Coefficient of MD (Mean) = MD(Mean) / Mean. Coefficient = 9.68 / 46 ≈ 0.2104.

Correct Answer: B. The most commonly used measure of dispersion, indicating the spread of data around the mean. Standard deviation is the primary and most widely used statistical tool to measure the variability or spread of observations in a distribution relative to their arithmetic mean.

Correct Answer: B. A high degree of consistency and homogeneity among observations. A small standard deviation implies that the data points tend to be close to the mean, indicating less variability and greater consistency or uniformity within the dataset. For example, test scores {75, 76, 77} have a smaller SD (indicating more consistency) than {60, 76, 92}, even though both sets have the same mean (76).

Correct Answer: B. A relative measure of dispersion used to compare variability between datasets with different means or units. CV is calculated as (Standard Deviation / Mean) * 100%. It expresses the standard deviation as a percentage of the mean, providing a unitless measure useful for comparing the relative spread of datasets. For example, comparing the variability of weights (in kg) and heights (in cm) requires CV.

Correct Answer: C. It is based on all observations and is rigidly defined. Standard Deviation utilizes every data point in its calculation, making it comprehensive. It is also rigidly defined, meaning its calculation method is unambiguous.

Correct Answer: D. It gives more weight to extreme items and is not easy to calculate or understand. Because SD involves squaring deviations, extreme values have a disproportionately large impact on the result. Additionally, its calculation is more complex than measures like Range or QD, and its interpretation might not be intuitive.

Correct Answer: B. The degree of distortion from a symmetrical (bell-shaped) distribution. Skewness quantifies the lack of symmetry in a dataset’s distribution. A perfectly symmetrical distribution has zero skewness.

Correct Answer: C. The bulk of observations are on the left side, and the right tail is longer; Mean > Median > Mode. In positive skewness, most data points cluster at the lower end, with the tail extending towards the higher values on the right. The mean is pulled towards the longer tail, making it greater than the median, which is greater than the mode. Example: Income distribution in many societies.

Correct Answer: A. The bulk of observations are on the right side, and the left tail is longer; Mean < Median < Mode. In negative skewness, most data points cluster at the higher end, with the tail extending towards the lower values on the left. The mean is pulled towards the longer left tail, making it less than the median, which is less than the mode. Example: Age at retirement.

Correct Answer: C. The peakedness or flatness of the distribution, related to the tails and outliers. Kurtosis describes the shape of the distribution’s peak and the weight of its tails compared to a normal distribution. It helps understand the presence of extreme values (outliers).

Correct Answer: C. It has a peak higher and sharper than a normal (Mesokurtic) distribution and fatter tails. Leptokurtic distributions are more peaked in the center and have heavier/fatter tails, meaning there’s a higher probability of extreme values compared to a normal distribution.

Correct Answer: A. It has a peak lower and flatter than a normal (Mesokurtic) distribution and thinner tails. Platykurtic distributions are less peaked (flatter) in the center and have thinner tails, indicating a lower probability of extreme values compared to a normal distribution.

Correct Answer: C. Mesokurtic. A mesokurtic distribution has a level of peakedness and tail weight that is considered ‘normal’ or standard, like the bell-shaped normal distribution.

Correct Answer: C. Symmetrical Distribution. A value of β₁ = 0 indicates that the third central moment (μ₃) is zero, which is characteristic of a symmetrical distribution.

Correct Answer: D. Leptokurtic Distribution. According to the specified criteria, β₂ > 0 corresponds to a leptokurtic distribution, which has a higher peak and fatter tails than a mesokurtic (normal-like) distribution.

Correct Answer: A. 3.03. Mean (x̄) = (2+3+7+8+10) / 5 = 30 / 5 = 6. Σx² = 2² + 3² + 7² + 8² + 10² = 4 + 9 + 49 + 64 + 100 = 226. Using formula SD = √{Σx²/n – (x̄)²} = √{226/5 – (6)²} = √{45.2 – 36} = √9.2 ≈ 3.03.

Correct Answer: C. 12%. CV = (Standard Deviation / Mean) * 100% = (6 / 50) * 100% = 0.12 * 100% = 12%.

Correct Answer: C. 7.21. Using the formula SD = √{Σfx²/N – (Σfx/N)²}. N = Σf = 100. Σfx = 3550. Σfx² = 131225. SD = √{131225/100 – (3550/100)²} = √{1312.25 – (35.5)²} = √{1312.25 – 1260.25} = √52 ≈ 7.21.

Correct Answer: A. 20.31%. Mean (x̄) = 35.5. SD (σ) = 7.21. CV = (σ / x̄) * 100% = (7.21 / 35.5) * 100% ≈ 0.2031 * 100% = 20.31%.

Correct Answer: C. Symmetrical. A value of β₁ = 0 indicates that the third central moment (μ₃) is zero, which is characteristic of a symmetrical distribution.

Correct Answer: A. Leptokurtic. First, calculate β₂ = μ₄ / μ₂² = 36000 / (120)² = 36000 / 14400 = 2.5. Since β₂ = 2.5 is greater than 0, according to the specified criteria (Leptokurtic: β₂ > 0), the distribution is classified as Leptokurtic.