TGPSC MPSO / ASO · Paper II · Topic 2
TGPSC MPSO / ASO Central Tendency, Dispersion, Skewness & Kurtosis questions with answers
35 questions from TGPSC ASO Paper II Mock 1 with answers and TGPSC ASO Paper II Challenge 1 with answers. Try each one, then tap Show answer.
2.1Arithmetic, geometric & harmonic meansNotesPractise this chapter · Set 1 free
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The mean of 20 observations was found to be 15. Later it was found that one observation, 52, had been recorded as 25, and that another observation, 30, should not have been included at all. What is the correct mean of the remaining observations?
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Answer B. 15.63
Correct total = 300 − 25 + 52 − 30 = 297 for 19 observations, so the mean = 297 / 19 ≈ 15.63.
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
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Consider the following statements about the geometric mean (GM) of positive values:
1. The logarithm of the GM equals the arithmetic mean of the logarithms of the observations.
2. The GM of the ratios x/y of paired values equals the ratio of the GM of x to the GM of y.
3. The GM of a set of values is never smaller than their arithmetic mean.
Which of the statements given above is/are correct?
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Answer A. 1 and 2 only
log GM = Σ log x / n by definition. Since Π(x/y) = Πx / Πy, the n-th root gives GM(x/y) = GM(x)/GM(y), which is why the GM suits ratios and index numbers. The GM is never larger than the AM, so statement 3 is false.
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What is the arithmetic mean of the squares of the first 10 natural numbers, 1², 2², …, 10²?
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Answer B. 38.5
Σk² = n(n + 1)(2n + 1)/6 = 385 for n = 10, so the mean is 385/10 = 38.5.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability, unit on measures of central tendency)
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Match List I (average) with List II (most suitable use):
List I:
A. Arithmetic mean
B. Geometric mean
C. Harmonic mean
D. Weighted arithmetic mean
List II:
1. Average speed when equal distances are covered at different speeds
2. Average rate of growth of population over several years
3. Average of items of unequal importance, such as a cost of living index
4. Average marks of students in a class
Select the correct answer using the code given below.
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Answer D. A-4, B-2, C-1, D-3
AM is the everyday average (marks); GM averages growth rates; HM averages speeds over equal distances; the weighted AM gives items importance weights, as in a cost of living index.
Source: Telugu Akademi B.Com/B.A. Statistics, Measures of Central Tendency (AM, GM, HM)
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The geometric mean of a set of positive observations is 6. If every observation is squared, what is the geometric mean of the new set?
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Answer D. 36
(Π x²)^(1/n) = ((Π x)^(1/n))² = 6² = 36.
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
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A car travels from A to B at 40 km/h and returns over the same distance at 60 km/h. What is its average speed for the whole journey?
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Answer D. 48
Equal distances are covered at different speeds, so the harmonic mean applies: 2 × 40 × 60 / (40 + 60) = 4800 / 100 = 48 km/h.
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Consider the following statements:
1. The geometric mean is the appropriate average for averaging rates of growth and ratios.
2. The harmonic mean is the appropriate average of speeds when equal distances are covered at different speeds.
Which of the statements given above is/are correct?
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Answer C. Both 1 and 2
GM suits relative changes such as growth rates and ratios (and is used in index numbers); HM suits rates such as speed when the numerator quantity (distance) is fixed.
Source: Telugu Akademi B.Com/B.A. Statistics, Measures of Central Tendency (AM, GM, HM)
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Consider the following statements about the arithmetic mean x̄ of a set of observations:
1. x̄ always lies between the smallest and the largest observation (both included).
2. The sum of the deviations Σ(x − x̄) is zero only when the distribution is symmetrical.
Which of the statements given above is/are correct?
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Answer A. 1 only
Only 1. The mean always lies between the extreme values. The sum of deviations from the mean is zero for all data, symmetrical or not.
Source: Standard definition (BRAOU Sem1 Descriptive Statistics & Probability, unit on measures of central tendency)
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A constant 5 is added to every one of a set of positive observations. Which of the following averages increases by exactly 5?
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Answer D. The arithmetic mean only
Only the arithmetic mean. For example, for 1 and 9 the GM is 3, but for 6 and 14 it is √84 ≈ 9.17, not 8.
Source: Standard definition (BRAOU Sem1 Descriptive Statistics & Probability, unit on measures of central tendency)
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A car covers the first 60 km of a journey at 20 km/h and the next 60 km at 30 km/h. It then travels for 4 hours at 60 km/h. What is its average speed for the whole journey?
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Answer C. 40 km/h
The average speed is 40 km/h. The first two parts take 60 ÷ 20 = 3 hours and 60 ÷ 30 = 2 hours, and the last part covers 4 × 60 = 240 km. Total distance = 60 + 60 + 240 = 360 km and total time = 3 + 2 + 4 = 9 hours. Average speed = 360 ÷ 9 = 40 km/h; neither the plain AM nor the plain HM of the speeds fits, because the parts are equal neither in distance nor in time.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability, measures of central tendency)
2.1Median, mode, quartiles, deciles & percentilesNotesPractise this chapter · Set 1 free
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For the distribution below, find the mean and the median, and then estimate the mode by the empirical relation.
Class: 0–10, 10–20, 20–30, 30–40, 40–50
Frequency: 4, 10, 16, 12, 8
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Answer B. 26.625
Mean = Σfx/N = 1350/50 = 27. Median: N/2 = 25 in 20–30, = 20 + (25 − 14)/16 × 10 = 26.875. Mode ≈ 3 × 26.875 − 2 × 27 = 26.625.
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
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Consider the following statements:
1. For a grouped distribution with unequal class intervals, the frequencies should be adjusted to a common class width before the mode is calculated by the formula.
2. For a grouped distribution with unequal class intervals, the classes must be made equal before the median can be calculated by the formula.
Which of the statements given above is/are correct?
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Answer A. 1 only
The interpolation formula for the mode compares the modal class frequency with those of the adjoining classes, so with unequal widths the frequencies (or frequency densities) must first be adjusted. The median formula L + (N/2 − cf)/f × h needs only cumulative frequencies and the width of the median class, so equal classes are not required.
Source: Telugu Akademi / APSCHE B.Com Business Statistics, Measures of Central Tendency
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Find the median daily wage (₹) of the following workers.
Daily wage: Below 200, 200–300, 300–400, 400–500, 500 and above
No. of workers: 10, 18, 32, 25, 15
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Answer C. 368.75
The open ends do not matter for the median. N = 100, N/2 = 50; c.f.: 10, 28, 60, 85, 100, so the median class is 300–400. Median = 300 + (50 − 28)/32 × 100 = ₹368.75.
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
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For the observations 2, 5, 9, 14 and 20, the sum Σ|x − a| is the least for which value of a?
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Answer C. 9
Σ|x − a| is least when a is the median. The median of 2, 5, 9, 14, 20 is 9 (Σ|x − 9| = 27, while Σ|x − 10| = 28).
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability, unit on median, mode and partition values)
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Which of the following are merits of the median?
1. It can be found for a distribution with open-end classes.
2. It can be located graphically.
3. It is based on all the observations.
4. It is well suited to further algebraic treatment.
Select the correct answer using the code given below.
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Answer D. 1 and 2 only
1 and 2. The median works with open-end classes and can be read from ogives, but it ignores most values and is not suited to algebra.
Source: Standard definition (BRAOU Sem1 Descriptive Statistics & Probability, unit on median, mode and partition values)
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Find the median of the following discrete frequency distribution.
x: 10, 20, 30, 40, 50, 60
f: 12, 9, 8, 10, 6, 5
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Answer D. 30
N = 50, (N + 1)/2 = 25.50. Cumulative frequencies: 12, 21, 29, 39, 45, 50. The first c.f. ≥ 25.50 is 29, against x = 30. Median = 30.
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
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Consider the following statements:
1. The second quartile, the fifth decile and the fiftieth percentile of a distribution are all equal to the median.
2. The seventh decile of a distribution is the same as its seventy-fifth percentile.
Which of the statements given above is/are correct?
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Answer A. 1 only
Quartiles divide the data into 4 parts, deciles into 10 and percentiles into 100. So Q2 = D5 = P50 = median. D7 cuts off 70% of the items, so D7 = P70; P75 equals Q3.
Source: Telugu Akademi / APSCHE B.Com Business Statistics, Measures of Central Tendency
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Match List I (measure for grouped data) with List II (formula). L = lower boundary, f = frequency and h = width of the class concerned; c = c.f. of the class before it; f1, f0, f2 = frequencies of the modal class and of the classes before and after it.
List I:
A. Mode
B. Median
C. Third quartile
D. Seventh decile
List II:
1. L + (7N/10 − c)/f × h
2. L + (f1 − f0)/(2f1 − f0 − f2) × h
3. L + (3N/4 − c)/f × h
4. L + (N/2 − c)/f × h
Select the correct answer using the code given below.
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Answer C. A-2, B-4, C-3, D-1
The mode uses the frequencies on either side of the modal class (2). Every partition value is L + (position − c)/f × h with its own position: median N/2 (4), Q3 3N/4 (3), D7 7N/10 (1).
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
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A moderately skewed distribution has mean 45, mode 36 and standard deviation 10. Consider the following statements:
1. The median is about 42.
2. The distribution is positively skewed.
3. Karl Pearson's coefficient of skewness is 0.9.
4. If 5 is added to every observation, the coefficient of skewness becomes 1.4.
Which of the statements given above are correct?
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Answer A. 1, 2 and 3 only
Statements 1, 2 and 3 are correct. From mode = 3 median − 2 mean, median = (2 × 45 + 36)/3 = 126/3 = 42. Mean > median > mode, so the skewness is positive, and Karl Pearson's coefficient = (45 − 36)/10 = 0.9. Adding 5 to every value raises the mean and the mode by 5 and leaves the SD unchanged, so the coefficient stays 0.9; statement 4 is false.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability, skewness, moments and kurtosis)
2.2Range, quartile deviation, mean deviation, SD & CVNotesPractise this chapter · Set 1 free
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The mean and standard deviation of 20 observations were found to be 40 and 5. It was later found that one observation 35 had been wrongly copied as 53. What is the correct standard deviation?
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Answer A. 4.12
Σx = 800 → corrected 800 − 53 + 35 = 782, mean = 39.1. Σx² = 20(5² + 40²) = 32500 → corrected 32500 − 2809 + 1225 = 30916. σ² = 30916/20 − 39.1² = 1545.8 − 1528.81 = 16.99, so σ ≈ 4.12.
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What is the coefficient of variation of 12, 15, 18, 21 and 24?
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Answer A. 23.57%
Mean = 18; squared deviations 36, 9, 0, 9, 36 sum to 90; σ = √(90/5) = √18 = 4.24; CV = 4.24/18 × 100 = 23.57%.
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
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Consider the following statements about measures of dispersion:
1. Absolute measures are expressed in the units of the data.
2. Relative measures are pure numbers and can compare series in different units.
3. The coefficient of variation is an absolute measure.
Which of the above are correct?
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Answer C. 1 and 2 only
Absolute measures (range, QD, MD, SD) carry the data's units; relative measures (coefficients, CV) are unit-free ratios. CV is relative, so 3 is false.
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
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A series has mean 20 and SD 4. Every observation is multiplied by 3 and then 5 is added. What is the coefficient of variation of the new series?
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Answer B. 18.46%
New mean = 3 × 20 + 5 = 65; new SD = 3 × 4 = 12; CV = 12/65 × 100 = 18.46%.
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
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Which of the following are relative measures of dispersion?
1. Coefficient of quartile deviation
2. Standard deviation
3. Coefficient of variation
4. Mean deviation
Select the correct answer using the code given below.
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Answer A. 1 and 3 only
Relative measures are pure numbers (coefficients). SD and MD are absolute measures expressed in the units of the data.
Source: NCERT Class 11 Statistics for Economics, Measures of Dispersion
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Consider the following statements:
1. Standard deviation is independent of change of origin.
2. Standard deviation is independent of change of scale.
Which of the statements given above is/are correct?
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Answer A. 1 only
If u = (x − A)/h, then σₓ = |h| σᵤ: shifting the origin A has no effect, but dividing or multiplying by h changes the SD.
Source: Gupta, S.C. & Kapoor, V.K., Fundamentals of Mathematical Statistics, Measures of Dispersion
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To compare the variability of the heights (in cm) and weights (in kg) of a group of students, the most suitable measure is the:
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Answer B. Coefficient of variation
Heights and weights are in different units. The coefficient of variation (σ/x̄ × 100) is a unit-free relative measure, so it can compare them.
Source: Gupta, S.C. & Kapoor, V.K., Fundamentals of Mathematical Statistics, Measures of Dispersion
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A variable X has mean 50 and standard deviation 10, and Y = (60 − 2X)/5. Consider the following statements:
I. The mean of Y is −8.
II. The standard deviation of Y is −4.
Which of the statements given above is/are correct?
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Answer A. Only I is correct
Only statement I is correct. Y = (60 − 2X)/5 = 12 − 0.4X, so mean of Y = 12 − 0.4 × 50 = −8. A change of scale multiplies the SD by the absolute value of the factor, so SD of Y = |−0.4| × 10 = 4. A standard deviation can never be negative, so statement II is false.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability, measures of dispersion)
2.2Skewness & kurtosisNotesPractise this chapter · Set 1 free
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Karl Pearson's coefficient of skewness of X is 0.8 and β₂ of X is 3.6. If Y = 5 − 2X, the coefficient of skewness and β₂ of Y are
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Answer B. Skewness changes sign to −0.8; β₂ stays 3.6
Origin changes cancel. Multiplying by −2 multiplies Mean − Mode by −2 and σ by |−2| = 2, so Sk changes sign: −0.8. β₂ = μ₄/μ₂² has the scale factor 2⁴ in both parts and stays 3.6.
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
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Assertion (A): Two distributions can have the same mean, the same standard deviation and the same skewness, and yet differ in shape.
Reason (R): Kurtosis, which depends on the fourth moment, describes peakedness that the first three moments do not fix.
Select the correct answer using the code given below.
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Answer A. Both A and R are true, and R is the correct explanation of A
Both are true and R explains A. Mean, SD and skewness use the first three moments; peakedness depends on μ4.
Source: Standard definition (BRAOU Sem1 Descriptive Statistics & Probability, unit on moments, skewness and kurtosis)
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For 10 observations Σx = 500 and Σx² = 25,160. The mode is 47. What is Karl Pearson's coefficient of skewness?
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Answer C. 0.75
x̄ = 500/10 = 50. σ² = 25160/10 − 50² = 16, so σ = 4. Sk = (Mean − Mode)/σ = 3/4 = 0.75.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability, unit on moments, skewness and kurtosis)
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Consider the following statements about Sheppard's corrections:
1. They apply to grouped continuous distributions whose frequencies taper off to zero at both ends.
2. The third central moment μ₃ needs no correction.
Which of the statements given above is/are correct?
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Answer C. Both 1 and 2
Sheppard's corrections remove the grouping error under exactly those conditions; μ₂ and μ₄ are corrected, while μ₁ and μ₃ need none because the errors cancel.
Source: Medha AI, based on BRAOU/APSCHE degree textbooks (ASO Paper II)
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Match List I (central moment) with List II (what it gives):
List I:
A. First central moment μ1
B. Second central moment μ2
C. Third central moment μ3
D. Fourth central moment μ4
List II:
1. Basis of kurtosis (β2)
2. Always zero
3. Variance
4. Basis of skewness (β1)
Select the correct answer using the code given below.
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Answer C. A-2, B-3, C-4, D-1
The first moment about the mean is always zero; μ2 is the variance; β1 = μ3² / μ2³ measures skewness; β2 = μ4 / μ2² measures kurtosis.
Source: S.C. Gupta & V.K. Kapoor, Fundamentals of Mathematical Statistics, Moments, Skewness and Kurtosis
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Match List I (value) with List II (shape of the distribution):
List I:
A. γ1 = 0
B. γ2 > 0
C. γ1 < 0
D. γ2 < 0
List II:
1. Platykurtic
2. Negatively skewed
3. Leptokurtic
4. Symmetrical
Select the correct answer using the code given below.
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Answer C. A-4, B-3, C-2, D-1
A-4, B-3, C-2, D-1. γ1 gives skewness (0 symmetrical, negative = negative skew); γ2 = β2 − 3 gives kurtosis (> 0 lepto, < 0 platy).
Source: Standard definition (BRAOU Sem1 Descriptive Statistics & Probability, unit on moments, skewness and kurtosis)
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A moderately skewed distribution has mean 60, median 66 and standard deviation 12. What are Karl Pearson's coefficient of skewness and the approximate mode?
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Answer A. Coefficient −1.5, mode 78
The coefficient is −1.5 and the mode is about 78. Karl Pearson's coefficient using the median is 3(mean − median)/SD = 3(60 − 66)/12 = −1.5. The empirical relation mode = 3 median − 2 mean gives 198 − 120 = 78. Mean < median < mode, which agrees with the negative skewness.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability, skewness, moments and kurtosis)
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Assertion (A): If β2 = 3 for a distribution, the distribution must be normal.
Reason (R): For a normal distribution, β2 = 3.
Choose the correct option.
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Answer D. A is false, but R is true
A is false and R is true. Every normal distribution has β2 = μ4/μ2² = 3, so R holds. But the converse does not follow: many non-normal distributions, including skewed ones, also have β2 = 3. β2 = 3 only tells us the peakedness is the same as that of the normal curve (mesokurtic).
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability, skewness, moments and kurtosis)
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