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TGPSC ASO Paper II Challenge 1 with answers
- 25 questions
- 30 minutes
- +1 right · 0 wrong
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Topics in this test (8)
Questions and answers
- Q1 Introduction to Statistics
In a histogram with unequal class widths, the rectangle over the class 0–10 has height 3 cm and stands for a frequency of 15. On the same scale, the rectangles over the classes 10–25 and 25–30 have heights 4 cm and 6 cm. What are the frequencies of the classes 10–25 and 25–30?
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Answer A. 30 and 15
The frequencies are 30 and 15. In a histogram the area of a rectangle, not its height, stands for the frequency. The rectangle over 0–10 has area 3 × 10 = 30 units for a frequency of 15, so 1 unit of area = 0.5. The class 10–25 has area 4 × 15 = 60, giving 30, and the class 25–30 has area 6 × 5 = 30, giving 15.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability, classification, tabulation and diagrams)
- Q2 Introduction to Statistics
Consider the following statements about the collection of data:
1. A schedule is filled in by the enumerator, whereas a questionnaire is filled in by the informant himself.
2. The mailed questionnaire method is most suitable when most of the informants are illiterate.
3. Data collected by an agency for its own enquiry are primary data for that agency but secondary data for another agency that later uses them.
4. Secondary data should be checked for reliability, suitability and adequacy before they are used.
Which of the statements given above are correct?
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Answer A. 1, 3 and 4 only
Statements 1, 3 and 4 are correct. A schedule is carried and filled in by a trained enumerator, while a questionnaire is answered by the informant himself. A mailed questionnaire needs informants who can read and write; for illiterate informants the schedule method is used, so statement 2 is wrong. Whether data are primary or secondary depends on who uses them, and secondary data must always be scrutinised before use.
Source: Standard concept (BRAOU Sem1 Descriptive Statistics & Probability, collection of data)
- Q3 Central Tendency, Dispersion, Skewness & Kurtosis
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); answer and every option computed in Python
- Q4 Central Tendency, Dispersion, Skewness & Kurtosis
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)
- Q5 Central Tendency, Dispersion, Skewness & Kurtosis
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)
- Q6 Central Tendency, Dispersion, Skewness & Kurtosis
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); answer and every option computed in Python
- Q7 Central Tendency, Dispersion, Skewness & Kurtosis
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)
- Q8 Probability
For two events A and B, P(A) = 1/2, P(B | A) = 1/3 and P(A | B) = 1/4. Consider the following statements:
1. P(B) = 2/3
2. A and B are independent.
3. P(A ∪ B) = 1
4. The probability that neither A nor B occurs is 1/6.
Which of the statements given above are correct?
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Answer C. 1 and 3 only
Statements 1 and 3 are correct. By the multiplication theorem, P(A ∩ B) = P(A)P(B | A) = 1/6, and then P(B) = P(A ∩ B)/P(A | B) = (1/6)/(1/4) = 2/3. P(A ∪ B) = 1/2 + 2/3 − 1/6 = 1, so the probability of neither event is 0, not 1/6. A and B are not independent, because P(A)P(B) = 1/3 differs from P(A ∩ B) = 1/6.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability, probability)
- Q9 Probability
For two events, P(A) = 0.3, P(B) = 0.5 and P(A | B) = 0.4. Consider the following statements:
I. P(A | B′) = 0.6, where B′ is the complement of B.
II. P(B | A) = 2/3.
Which of the statements given above is/are correct?
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Answer B. Only II is correct
Only statement II is correct. P(A ∩ B) = P(A | B)P(B) = 0.4 × 0.5 = 0.2, so P(B | A) = 0.2/0.3 = 2/3. P(A ∩ B′) = 0.3 − 0.2 = 0.1 and P(B′) = 0.5, so P(A | B′) = 0.1/0.5 = 0.2, not 0.6. The value 0.6 is 1 − P(A | B), which is P(A′ | B), a different event.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability, probability)
- Q10 Probability
A box has two coins: one fair and one biased that shows heads with probability 3/4. A coin is picked at random and tossed three times, giving two heads and one tail. What is the probability that the biased coin was picked?
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Answer B. 9/17
The probability is 9/17. By the binomial rule, P(2 heads in 3 | fair) = 3 × (1/2)³ = 24/64 and P(2 heads in 3 | biased) = 3 × (3/4)² × (1/4) = 27/64. With equal priors, P(biased | data) = 27/64 ÷ (27/64 + 24/64) = 27/51 = 9/17. The tail is evidence for the fair coin, so the posterior is only a little above 1/2.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability, probability); answer and every option computed in Python
- Q11 Probability
Two different numbers are chosen at random from 1, 2, 3, …, 9. What is the probability that their sum is even?
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Answer B. 4/9
The probability is 4/9. There are C(9, 2) = 36 equally likely pairs. The sum is even when both numbers are odd, C(5, 2) = 10 ways, or both are even, C(4, 2) = 6 ways. So the probability is 16/36 = 4/9.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability, probability); answer and every option computed in Python
- Q12 Random Variables, Distributions & Inference
The number of defects per sheet of cloth follows a Poisson distribution with mean 1. Three sheets are inspected, independently of one another. What is the probability that exactly one of the three sheets is free of defects?
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Answer A. 3e^−1(1 − e^−1)²
The probability is 3e^−1(1 − e^−1)². A sheet is free of defects with probability P(X = 0) = e^−1. The number of defect-free sheets among three is then binomial with n = 3 and p = e^−1. So P(exactly one) = C(3, 1) e^−1 (1 − e^−1)², about 0.44.
Source: Standard properties of the Poisson and binomial distributions (BRAOU Sem3 Discrete Distributions); answer and every option computed in Python (checker rewrite)
- Q13 Random Variables, Distributions & Inference
Calls reach desk A as a Poisson variable with mean 1 per hour and desk B as an independent Poisson variable with mean 2 per hour. What is the probability that the two desks together receive exactly 2 calls in an hour? (Take e^−3 = 0.0498.)
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Answer C. 0.2241
The probability is about 0.2241. The sum of independent Poisson variables is Poisson with mean 1 + 2 = 3. So P(total = 2) = e^−3 × 3²/2! = 0.0498 × 4.5 ≈ 0.2241. The same answer comes from adding the cases (0, 2), (1, 1) and (2, 0): e^−3(2 + 2 + 0.5).
Source: Standard method (BRAOU Sem3 Discrete Distributions); answer and every option computed in Python
- Q14 Random Variables, Distributions & Inference
Misprints per page in book A follow a Poisson distribution with mean 1, and in book B a Poisson distribution with mean 2. One of the two books is chosen at random and a page is opened at random. The page has no misprint. What is the probability that the book chosen was A?
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Answer C. e/(1 + e)
The probability is e/(1 + e), about 0.73. P(no misprint | A) = e^−1 and P(no misprint | B) = e^−2. By Bayes' theorem with equal priors, P(A | no misprint) = e^−1/(e^−1 + e^−2). Multiplying top and bottom by e², this is e/(e + 1).
Source: Poisson distribution with Bayes' theorem (BRAOU Sem3 Discrete Distributions; Sem1 Probability); answer and every option computed in Python (checker rewrite)
- Q15 Random Variables, Distributions & Inference
Consider the following statements:
I. If X is normal with mean μ and variance σ², then for constants a ≠ 0 and b, aX + b is normal with mean aμ + b and variance a²σ².
II. If X is normal with mean 50 and variance 16, then 2X has the same distribution as X1 + X2, where X1 and X2 are independent, each normal with mean 50 and variance 16.
Which of the statements given above is/are correct?
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Answer A. Only I is correct
Only statement I is correct. A linear function of a normal variable is normal, with mean aμ + b and variance a²σ². 2X has mean 100 and variance 2² × 16 = 64. X1 + X2 also has mean 100, but its variance is 16 + 16 = 32, because independent variances add. So the two distributions differ, and statement II is false.
Source: Standard method (BRAOU Sem4 Continuous Distributions, normal distribution)
- Q16 Random Variables, Distributions & Inference
A 95% confidence interval for a population mean μ, based on a random sample of 100 and the known population SD σ, is (48.08, 55.92), using z = 1.96. What is σ, and what is the decision on H0: μ = 50 against H1: μ ≠ 50 at the 5% level?
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Answer D. σ = 20; do not reject H0
σ = 20 and H0 is not rejected. The interval is x̄ ± 1.96σ/√n, so x̄ = 52 and the half-width 3.92 = 1.96 × σ/10 gives σ = 20. The two-tailed 5% test of μ = 50 rejects H0 exactly when 50 lies outside the 95% interval, and 50 lies inside (48.08, 55.92). Directly, z = (52 − 50)/(20/10) = 1 < 1.96.
Source: Standard concept (BRAOU Sem6 Estimation & Testing of Hypothesis); answer and every option computed in Python
- Q17 Correlation & Regression
The two regression lines of a bivariate distribution are 3x + 2y = 26 and 6x + y = 31. What are the means of X and Y and the correlation coefficient?
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Answer A. x̄ = 4, ȳ = 7, r = −0.5
x̄ = 4, ȳ = 7 and r = −0.5. Both lines pass through (x̄, ȳ), and solving them together gives x = 4, y = 7. Taking the first as Y on X gives byx = −3/2, and the second as X on Y gives bxy = −1/6; their product 1/4 is at most 1, so this choice is valid (the other choice gives a product of 4, which is impossible). r = −√(1/4) = −0.5, negative because both coefficients are negative.
Source: Standard method (BRAOU Sem5 Correlation, Regression & Sampling, regression); answer and every option computed in Python
- Q18 Correlation & Regression
For 5 pairs of observations, Σx = 20, Σy = 40, Σxy = 180 and Σx² = 100. Using the regression line of Y on X, estimate Y when X = 10.
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Answer D. 14
The estimate is 14. byx = (nΣxy − ΣxΣy)/(nΣx² − (Σx)²) = (900 − 800)/(500 − 400) = 1. With x̄ = 4 and ȳ = 8, the line is y − 8 = 1 × (x − 4), that is y = x + 4. At x = 10, y = 14.
Source: Standard method (BRAOU Sem5 Correlation, Regression & Sampling, regression); answer and every option computed in Python
- Q19 Correlation & Regression
Consider the following statements:
I. If U = 3 − 2X and V = 5 − 4Y, then r(U, V) = r(X, Y).
II. If every value of X is multiplied by −3 and Y is left unchanged, the correlation coefficient changes sign but not size.
Which of the statements given above is/are correct?
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Answer C. Both I and II are correct
Both statements are correct. The correlation coefficient is unaffected by a change of origin, and a change of scale affects only its sign: r(U, V) = (bd/|bd|) r(X, Y) for U = a + bX, V = c + dY. In statement I, b = −2 and d = −4 have the same sign, so r(U, V) = r(X, Y). In statement II, only X is multiplied by a negative number, so r becomes −r.
Source: Standard method (BRAOU Sem5 Correlation, Regression & Sampling, correlation)
- Q20 Time Series & Index Numbers
In a cost of living index by the family budget method, the groups food, rent, clothing, fuel and others have weights 50, 20, 10, 10 and 10 and current price indices 180, 150, 200, 120 and 140. A family earned 6,000 rupees a month in the base period. What are the cost of living index and the income needed now to keep the same standard of living?
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Answer D. 166 and 9,960 rupees
The index is 166 and the income needed is 9,960 rupees. CLI = Σ(IW)/ΣW = (9,000 + 3,000 + 2,000 + 1,200 + 1,400)/100 = 16,600/100 = 166. To keep the base-period standard, income must rise in the same proportion as the index. Required income = 6,000 × 166/100 = 9,960 rupees.
Source: Standard method (degree Applied Statistics: index numbers); answer and every option computed in Python
- Q21 Time Series & Index Numbers
A series for seven years reads 12, 15, 18, 14, 20, 22, 19. Match List I with List II.
List I:
- A. 3-yearly moving average for year 2
- B. 3-yearly moving average for year 3
- C. First 4-yearly moving average before centring
- D. 4-yearly centred moving average for year 3
List II:
- 1. 15.75
- 2. 15
- 3. 14.75
- 4. 15.67
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Answer A. A-2, B-4, C-3, D-1
The correct match is A-2, B-4, C-3, D-1. The 3-yearly averages are (12 + 15 + 18)/3 = 15 for year 2 and (15 + 18 + 14)/3 = 15.67 for year 3. The first two 4-yearly averages are 59/4 = 14.75 and 67/4 = 16.75; they fall between years 2–3 and 3–4. Centring them gives (14.75 + 16.75)/2 = 15.75 for year 3.
Source: Standard method (degree Applied Statistics: time series analysis)
- Q22 Time Series & Index Numbers
Under the additive model, the preliminary seasonal effects of the four quarters are +18, −10, +6 and −22; they must be adjusted so that they add up to zero. The quarterly trend is Yc = 300 + 4X, with X in quarters and X = 0 at the first quarter of year 1. What is the forecast for the third quarter of year 4?
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Answer C. 364
The forecast is 364. The preliminary effects add up to −8, so their mean is −2; subtracting −2 (adding 2) from each gives +20, −8, +8 and −20, which add up to zero. The third quarter of year 4 is X = 3 × 4 + 2 = 14, so the trend value is 300 + 4 × 14 = 356. Under the additive model the forecast is 356 + 8 = 364.
Source: Standard method (degree Applied Statistics: time series, additive model and seasonal effects); answer and every option computed in Python
- Q23 Sampling Techniques
A pilot survey suggests that about 20% of the 2,000 households in a mandal own a tractor. The proportion is to be estimated within ±0.04 with about 95% confidence (use z = 2). Using n0 = z²pq/d² and the finite population correction n = n0/(1 + n0/N), rounded up, what sample size is needed?
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Answer B. 334
About 334 households are needed. With p = 0.2 and q = 0.8, n0 = 4 × 0.16/0.0016 = 400. Correcting for N = 2,000, n = 400/(1 + 400/2000) = 400/1.2 ≈ 333.3, rounded up to 334.
Source: Standard method (BRAOU Sem5 Correlation, Regression & Sampling, sampling theory); answer and every option computed in Python
- Q24 Basics of Computers
A memory chip is organised as 32K × 8 bits (32K locations of one byte each). How many address lines does one chip need, and how many such chips are needed to build a 256 KB byte-addressable memory?
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Answer B. 15 address lines; 8 chips
One chip needs 15 address lines, and 8 chips are needed. 32K = 32 × 1,024 = 2¹⁵ locations, so 15 address lines are needed to select one of them. Each chip stores 32 KB, because each location holds 8 bits = 1 byte. So 256 KB ÷ 32 KB = 8 chips.
Source: Standard computer fundamentals (number systems, data representation and logic operations); answer and every option computed in Python
- Q25 Basics of Computers
In MS-Excel, cell C2 contains the formula =$A2*B$1. This cell is copied and pasted into cell E5. Which formula will E5 contain?
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Answer D. =$A5*D$1
E5 will contain =$A5*D$1. Copying from C2 to E5 moves the formula 2 columns right and 3 rows down. In $A2 the column is fixed but the row is relative, so it becomes $A5. In B$1 the row is fixed but the column is relative, so it becomes D$1.
Source: Standard MS-Office knowledge (Excel cell references)
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