NDA Statistics, Correlation and Regression previous year questions with solutions

Answer: (b) 39

Method. reading a cumulative frequency inside a class by linear interpolation (ogive)

Cumulative frequencies at the upper boundaries: 162 → 12, 164 → 12 + 15 = 27, 166 → 27 + 24 = 51, 168 → 64.
165 cm lies inside the class 164–⁠166 (width 2, frequency 24); with the students spread uniformly through a class, the part from 164 to 165 holds (1/⁠2) × 24 = 12.
Number with height ≤ 165 cm ≈ 27 + 12 = 39.

Answer: (c) 164.41 cm

Method. median of grouped data, L + N/⁠2 − cff × h

N = 12 + 15 + 24 + 13 = 64, N/⁠2 = 32.
Cumulative frequencies: 12, 27, 51, 64, so the median class is 164–⁠166.
L = 164, cf = 27, f = 24, h = 2.
Median = 164 + 32 − 2724 × 2 = 164 + 512 = 164.4167 cm ≈ 164.41 cm (option (c)).

Answer: (d) 164.9 cm

Method. mode of grouped data, L + f₁ - f₀2f₁ - f₀ - f₂ × h

The height occurring most often is the mode. Modal class (largest frequency 24): 164–⁠166.
L = 164, f₁ = 24, f₀ = 15, f₂ = 13, h = 2.
Mode = 164 + 24 − 1548 − 15 − 13 × 2 = 164 + 920 × 2 = 164 + 0.9 = 164.9 cm.

Answer: (c) histogram

Method. continuous (class-interval) frequency data is drawn as a histogram

Height is a continuous variable and the data are in adjoining class intervals 160–⁠162, 162–⁠164, …
Such a distribution is shown by a histogram: adjacent rectangles on the class intervals with areas proportional to the frequencies.
Bar charts and pie charts suit discrete categories, not continuous classes.

Answer: (d) 36

Method. write each angle in terms of a common value k and use total 360°

Let 9p = 3q = 2r = 6s = k: p = k/⁠9, q = k/⁠3, r = k/⁠2, s = k/⁠6.
p + q + r + s = k(2 + 6 + 9 + 318) = 10k9 = 360 ⇒ k = 324.
p = 36, q = 108.
4p − q = 144 − 108 = 36.

Answer: (b) 72°

Method. pie-chart angle = (part/⁠total) × 360°

Total = 30 + 70 + 50 = 150.
Angle for Science = 30150 × 360° = 72°.

Answer: (b) the area of the rectangle

Method. histogram with unequal class widths

With unequal widths the height is frequency density (frequency ÷ width).
Then area = height × width = frequency, so frequency is proportional to the area of the rectangle.

Answer: (c) 40 − 50

Method. compare absolute differences of frequencies class by class

Differences |Physics − Mathematics|: 10-20: 2; 20-30: 10; 30-40: 8; 40-50: 11; 50-60: 5; 60-70: 4.
The largest, 11, is in 40-50.

Answer: (a) 1 only

Method. modal class by highest frequency; medians by interpolation in the median class

1. Physics frequencies 8, 11, 30, 26, 15, 10: highest is 30, in 30-40. Mode = 30 + 10 × 30 − 112(30) − 11 − 26 = 30 + 19023 ≈ 38.3, in 30-40. True.
2. Both subjects have N = 100, so N/⁠2 = 50.
Physics cf: 8, 19, 49, 75, …; median class 40-50: median = 40 + 10 × 50 − 4926 ≈ 40.4.
Mathematics cf: 10, 31, 69, …; median class 30-40: median = 30 + 10 × 50 − 3138 = 35.
Physics median (≈ 40.4) is greater than Mathematics (35), so 2 is false.
Only 1 is correct.

Answer: (c) 40.9

Method. mean of grouped data using class marks

Class marks 15, 25, 35, 45, 55, 65; Physics frequencies 8, 11, 30, 26, 15, 10 (N = 100).
Σ fx = 120 + 275 + 1050 + 1170 + 825 + 650 = 4090.
Mean = 4090100 = 40.9.

Answer: (b) 2.5-7.5, 7.5-12.5, 12.5-17.5, 17.5-22.5, 22.5-27.5

Method. class limits lie half a class width either side of the class mark

Consecutive class marks differ by 5, so the class width is 5.
Each class runs from (mark − 2.5) to (mark + 2.5).
First five classes: 2.5-7.5, 7.5-12.5, 12.5-17.5, 17.5-22.5, 22.5-27.5.

Answer: (b) Median

Method. an ogive is a cumulative frequency curve; its N/⁠2 point gives the median

An ogive plots cumulative frequency against the class boundary.
Draw the horizontal line at cumulative frequency N2; the x-coordinate where it meets the ogive is the median.
(The point where the less-than and more-than ogives intersect also gives the median.) An ogive gives positional values only: median, quartiles and percentiles. The mode is read from a histogram or frequency curve, and the mean and geometric mean are computed from the data, not read off a cumulative curve.

Answer: (c) 72°

Method. sector angle = percentage × 360°

Angle = 20100 × 360° = 72°.

Answer: (b) 10 : 9

Method. radii of pie diagrams ∝ ⁠√(total)⁠

Total of Family A = 3500 + 500 + 1500 + 2000 + 2500 = 10000.
Total of Family B = 2700 + 800 + 1000 + 1800 + 1800 = 8100.
Circle areas are proportional to the totals, so radii ∝ ⁠√(total)⁠.
r_A : r_B = ⁠√(10000)⁠: ⁠√(8100)⁠ = 100 : 90 = 10 : 9.

Answer: (c) 144°

Method. sector angle = percentage × 360°

Angle = 40100 × 360° = 144°.

Answer: (d) 1, 2 and 3

Method. forms of presentation of data

Data are presented textually (in descriptive sentences), in tables, and graphically (diagrams, charts, graphs).
All three are correct forms.

Answer: (a) Median

Method. intersection of the two ogives

At the intersection the 'less than' cumulative frequency equals the 'more than' cumulative frequency, each being N/⁠2.
The value of the variable with N/⁠2 observations on either side is the median, so the abscissa of P is the median.

Answer: (b) 25

Method. mean of grouped data Σf x/⁠Σf with class mid-points

Mid-points: 10, 20, 30, 40; frequencies 20, 30, 30, 20; Σf = 100.
Σf x = 200 + 600 + 900 + 800 = 2500.
Mean = 2500/⁠100 = 25.

Shortcut. The distribution is symmetric about 25, so the mean is 25.

Answer: (c) 137.5

Method. sum of squares Σk² = n(n + 1)(2n + 1)/⁠6

8² + … + 15² = 15Σ1k² - 7Σ1k² = 15.16.316 − 7.8.156 = 1240 − 140 = 1100.
There are 15 − 8 + 1 = 8 terms.
Mean = 1100/⁠8 = 137.5.

Answer: (b) 19

Method. sum of squares of the first n natural numbers

Mean = (1² + 2² + … + n²)/⁠n = n(n + 1)(2n + 1)6n = (n + 1)(2n + 1)6.
(n + 1)(2n + 1)/⁠6 = 130 ⇒ (n + 1)(2n + 1) = 780.
n = 19: 20 × 39 = 780 (holds). So n = 19.

Shortcut. Test the options: only n = 19 gives (n + 1)(2n + 1) = 780.

Answer: (d) nx̅ − x_n + kn

Method. new total = old total − removed value + added value

Old total = nx̅.
Replacing x_n by k: new total = nx̅ − x_n + k; the count stays n.
New mean = nx̅ − x_n + kn.

Answer: (b) 2ⁿ⁺¹ - n − 22^n-1

Method. arithmetico-geometric series S = Σ i(1/⁠2)^i-1

S = 1 + 22 + 32² + … + n2^n-1.
S2 = 12 + 22² + … + n2ⁿ.
Subtract: S2 = 1 + 12 + … + 12^n-1 − n2ⁿ = 2 − 22ⁿ − n2ⁿ.
S = 4 − n + 22^n-1 = 2ⁿ⁺¹ - n − 22^n-1.
Check n = 2: 1 + 2 × 12 = 2 and 8 − 42 = 2.

Answer: (c) 2ⁿ⁺¹ - n − 22ⁿ - 1

Method. mean = Σx_if_i / Σf_i; Σf_i is a GP

Σf_i = 1 + 12 + … + 12^n-1 = 2 − 12^n-1 = 2ⁿ - 12^n-1.
From the previous item Σx_if_i = 2ⁿ⁺¹ - n − 22^n-1.
Mean = 2ⁿ⁺¹ - n − 22^n-1 × 2^n-12ⁿ - 1 = 2ⁿ⁺¹ - n − 22ⁿ - 1.
Check n = 2: mean = 2/⁠(3/⁠2) = 4/⁠3 and 8 − 43 = 43.

Answer: (a) 10

Method. Σ(x − A) = Σx − nA

p = Σx − 10n and q = Σx − 20n.
p − q = 10n.
(10n)² = 10000 ⇒ 100n² = 10000 ⇒ n = 10.

Answer: (c) 5.5

Method. Σ⁹C_k = 2⁹ and Σk·⁹C_k = 9.2⁸

The number k + 1 has frequency ⁹C_k, k = 0 to 9.
Σf = Σ⁹C_k = 2⁹ = 512.
Σfx = Σ(k + 1)⁹C_k = 9.2⁸ + 2⁹ = 2⁸(9 + 2) = 11 × 256.
Mean = 11 × 256512 = 5.5.

Shortcut. Frequencies are symmetric (⁹C_k = ⁹C_9-k) about the middle of 1 to 10, so mean = (1 + 10)/⁠2 = 5.5.

Answer: (a) 10

Method. sort the data, then take means of the two parts

Sorted: 1, 1, 1, 2, 3, 3, 4, 4, 4, 5, 6, 6.
Lowest 8: 1 + 1 + 1 + 2 + 3 + 3 + 4 + 4 = 19, so m = 19/⁠8.
Highest 4: 4 + 5 + 6 + 6 = 21, so M = 21/⁠4.
2m + M = 19/⁠4 + 21/⁠4 = 40/⁠4 = 10.

Answer: (c) 27

Method. Σk² = n(n + 1)(2n + 1)/⁠6 and Σk = n(n + 1)/⁠2

x = 1n × n(n + 1)(2n + 1)6 = (n + 1)(2n + 1)6.
y = (n + 12)² = (n + 1)²4.
xy = 4(2n + 1)6(n + 1) = 2(2n + 1)3(n + 1) = 5542.
84(2n + 1) = 165(n + 1) ⇒ 168n + 84 = 165n + 165 ⇒ 3n = 81 ⇒ n = 27.

Answer: (d) 99

Method. Σ(x_i - a) = Σx_i - na

Σx_i - 100n = -20 and Σx_i - 92n = 140.
Subtract: 8n = 160 ⇒ n = 20; then Σx_i = 2000 − 20 = 1980.
Mean = 1980/⁠20 = 99.

Answer: (a) 0

Method. sum of deviations from the mean is zero

The mean is 99 (previous item), so Σ(x_i - 99) = 1980 − 20 × 99 = 0.

Answer: (c) 90

Method. Σ(x_i - y) = Σx_i - ny

1980 − 20y = 180 ⇒ 20y = 1800 ⇒ y = 90.

Answer: (d) 79

Method. mean of an AP with an odd number of terms = middle term

51 terms, a = 4, d = 3.
Mean = first + last2; last = 4 + 50 × 3 = 154, so mean = 4 + 1542 = 79.

Answer: (b) 79

Method. median of 51 ordered values = 26th value

The AP is increasing, so the median is the 26th term = 4 + 25 × 3 = 79.

Answer: (b) 0

Method. median equals mean here, and deviations from the mean sum to zero

For the AP, median = mean = 79.
Σ(x_i - 79) = Σx_i - 51 × 79 = 51 × 79 − 51 × 79 = 0.

Answer: (c) 27

Method. total frequency and mean of grouped data give two equations in p and q

Total frequency: 17 + (p + q) + 32 + (p − 3q) + 19 = 120 ⇒ 2p − 2q = 52 ⇒ p − q = 26.
Class marks: 10, 30, 50, 70, 90.
Σfx = 170 + 30(p + q) + 1600 + 70(p − 3q) + 1710 = 3480 + 100p − 180q.
Mean 50 ⇒ Σfx = 50 × 120 = 6000 ⇒ 100p − 180q = 2520 ⇒ 5p − 9q = 126.
Put p = q + 26: 5q + 130 − 9q = 126 ⇒ q = 1, so p = 27.

Answer: (a) 1

Method. same pair of equations as the previous item

From the previous item: p − q = 26 and 5p − 9q = 126.
Substituting p = q + 26: 5q + 130 − 9q = 126 ⇒ 4q = 4 ⇒ q = 1.

Answer: (b) 50

Method. mean is unchanged when every frequency is multiplied by the same constant

New mean = Σ(2f)x / Σ(2f) = 2Σfx / 2Σf = Σfx / Σf.
So the mean stays 50.

Answer: (c) 39.5

Method. mean of consecutive integers = (first + last)/⁠2

The natural numbers 15, 16, …, 64 form an AP.
Mean = 15 + 642 = 39.5.

Answer: (c) 0

Method. sum of deviations from the mean is zero

Σ(x_i - x̅) = Σx_i - n x̅ = n x̅ − n x̅ = 0, for any data.
(Here mean = 475/⁠5 = 95; deviations -22, -10, -3, 10, 25 add to 0.)

Answer: (b) 125

Method. total = mean × number

Sum of 12 observations = 12 × 75 = 900.
Sum of remaining 10 = 10 × 65 = 650.
Sum of the 2 discarded = 900 − 650 = 250, so their mean = 250/⁠2 = 125.

Answer: (c) reduced by 103

Method. a wrong tens digit changes the value by 10 × (digit error)

Tens digit recorded as 8 instead of 3: recorded value is 10(8 − 3) = 50 too large.
Correcting reduces the total by 50, so the average falls by 50/⁠15 = 103.

Answer: (b) 3

Method. weighted arithmetic mean

4x + 9(x − 1)x + (x − 1) = 6.
13x − 9 = 6(2x − 1) = 12x − 6.
x = 3.

Answer: (d) 100

Method. sum of deviations from A is Σx − nA

Σx − 2.5n = 50 and Σx − 3.5n = -50.
Subtracting: n(3.5 −⁠ 2.5) = 100, so n = 100.

Answer: (d) 4M3

Method. combined mean of two data sets

Combined mean = n(2M) + 2n(M)n + 2n = 4nM3n = 4M3.

Answer: (b) 39.7

Method. correct the total, then divide

Wrong total = 100 × 40 = 4000.
Correct total = 4000 − 83 + 53 = 3970.
Correct mean = 3970100 = 39.7.

Answer: (b) 50 and 100

Method. combined mean of two groups

Let b boys and g girls: b + g = 150 and 70b + 55g = 60 × 150 = 9000.
70b + 55(150 − b) = 9000, so 15b = 750, b = 50.
g = 100. Boys 50, girls 100.

Shortcut. By alligation, boys: girls = (60 − 55): (70 − 60) = 1 : 2, i.e. 50 and 100.

Answer: (d) 80, 20

Method. combined mean of two groups

Let the fraction of men be p; women 1 − p.
26p + 21(1 − p) = 25, so 5p = 4 and p = 0.8.
Men 80%, women 20%.

Shortcut. Men: women = (25 − 21): (26 − 25) = 4 : 1, i.e. 80% and 20%.

Answer: (c) Both 1 and 2

Method. basic properties of deviations from mean and median

(1) Σ(x_i - x̅) = Σx_i - nx̅ = 0: true.
(2) Σ|x_i - A| is minimum when A is the median: true.
Both are correct.

Answer: (d) n2

Method. Σ k ⁿC_k = n 2^n-1 and Σ ⁿC_k = 2ⁿ

Total frequency = nΣk=0 ⁿC_k = 2ⁿ.
Σ k ⁿC_k = Σ n ^n-1C_k-1 = n 2^n-1 (using k ⁿC_k = n ^n-1C_k-1).
Mean = n 2^n-12ⁿ = n2.

Shortcut. The frequencies are symmetric about the middle value (ⁿC_k = ⁿC_n-k), so the mean is the midpoint of 0 and n, i.e. n/⁠2.

Answer: (b) 20

Method. total of the observations, then remove the omitted ones

Sum of all 100 observations = 100 × 20 = 2000.
Sum of the four omitted = 21 + 21 + 18 + 20 = 80.
Remaining 96 observations: sum = 2000 − 80 = 1920.
Mean = 192096 = 20.

Shortcut. The omitted four average 80/⁠4 = 20, equal to the overall mean, so the mean is unchanged.

Answer: (a) 50

Method. combined mean of two groups

Let the number of boys be b; girls = 150 − b.
70b + 55(150 − b) = 150 × 60 = 9000.
15b + 8250 = 9000, so 15b = 750 and b = 50.

Shortcut. Boys: girls = (60 − 55): (70 − 60) = 1 : 2, so boys = 150/⁠3 = 50.

Answer: (b) The median remains the same but the mean will decrease

Method. effect of correcting one value on mean and median

Original sum = 5 × 32 = 160; corrected sum = 160 − 40 + 35 = 155, so mean = 31: the mean decreases.
Arranged in order, the median is the 3rd value, 33. Since 40 > 33, it is one of the two values above the median.
The corrected value 35 is also above 33, so the order of the middle value is unaffected: the median stays 33.

Answer: (a) 3150/⁠19

Method. harmonic mean n/⁠Σ(1/⁠x_i) with ⁿC_r = ⁿC_n-r

C(10, 3) = C(10, 7) = 120, C(10, 4) = C(10, 6) = 210, C(10, 5) = 252.
Σ1/⁠x = 2/⁠120 + 2/⁠210 + 1/⁠252 = (42 + 24 + 10)/⁠2520 = 76/⁠2520.
HM = 5 ÷ 762520 = 1260076 = 315019.

Answer: (a) 210

Method. equal money spent each year ⇒ average price is the harmonic mean

With ₹A spent each year, litres bought = A/⁠150 + A/⁠200 + A/⁠250 + A/⁠300.
Average price = 4AA(1/⁠150 + 1/⁠200 + 1/⁠250 + 1/⁠300) = 4(20 + 15 + 12 + 10)/⁠3000 = 1200057.
12000/⁠57 ≈ 210.5 ≈ 210.

Answer: (c) 1714

Method. HM of three numbers = 3 / Σ(1/⁠x)

¹⁰C₄ = 210, ¹⁰C₅ = 252, ¹⁰C₆ = 210.
1210 + 1252 + 1210 = 1105 + 1252 = 12 + 51260 = 171260.
H = 3 × 126017 = 378017.
270H = 270 × 173780 = 1714.

Answer: (d) 1

Method. mode = most frequent value

Counts: 1 → 4 times, 2 → 2, 4 → 2, 5 → 1, 6 → 1.
Mode = 1.

Answer: (c) 3

Method. median of a discrete frequency distribution via cumulative frequency

N = 4 + 6 + 9 + 7 = 26; median is the mean of the 13th and 14th values.
Cumulative frequencies: 4, 10, 19, 26.
The 11th to 19th values are all 3, so the 13th and 14th are 3.
Median = 3.

Answer: (d) 0

Method. sort each group of five and take the middle value

First five: -1, 1, 4, 3, 8 → sorted -1, 1, 3, 4, 8; M = 3.
Last five: 12, 17, 19, 9, 11 → sorted 9, 11, 12, 17, 19; N = 12.
4M − N = 12 − 12 = 0.

Answer: (d) 14

Method. express P, Q, R through a common value k

Let 5P = 4Q = R/⁠2 = k: P = k/⁠5, Q = k/⁠4, R = 2k.
P + Q = k5 + k4 = 9k20.
2P + 0.7R = 2k5 + 1.4k = 9k5.
Ratio = 9k/⁠209k/⁠5 = 14.

Answer: (d) n

Method. G = (product)^1/⁠n and 0 + 1 + … + (n − 1) = n(n −⁠ 1)/⁠2

Product = 2^{0 + 1 + … + (n − 1)} = 2^{n(n −⁠ 1)/⁠2}.
G = 2^{(n − 1)/⁠2}, so log₂G = n − 12.
1 + 2log₂G = 1 + (n − 1) = n.

Answer: (b) 2 − 12^n-1

Method. H = n / Σ(1/⁠x_i); the reciprocals form a GP

Σ1x_i = 1 + 12 + … + 12^n-1 = 1 − (1/⁠2)ⁿ1 − 1/⁠2 = 2 − 12^n-1.
H = nΣ(1/⁠x_i), so nH = Σ1x_i = 2 − 12^n-1.

Answer: (c) 2Q

Method. sum of squared deviations is least about the mean

S = Σ(y_i - a)² with y_i = 2x_i; this is least when a = mean of the y_i.
Mean of 2x_i = 2 × (mean of x_i) = 2Q.
(Directly: dS/⁠da = -2Σ(2x_i - a) = 0 ⇒ na = 2Σx_i ⇒ a = 2Q.)

Answer: (c) 35

Method. median of grouped data: l + N/⁠2 − cff × h

N = 1 + 2 + 4 + 6 + 4 + 3 = 20, so N/⁠2 = 10.
Cumulative frequencies: 1, 3, 7, 13, 17, 20; the 10th value lies in 30-40.
l = 30, cf = 7, f = 6, h = 10.
Median = 30 + 10 − 76 × 10 = 30 + 5 = 35.

Answer: (d) 10.5

Method. mean deviation about median = Σf|x − M| / N using class marks

Class marks x: 5, 15, 25, 35, 45, 55; median M = 35.
|x − 35|: 30, 20, 10, 0, 10, 20.
f|x − 35|: 30, 40, 40, 0, 40, 60; total 210.
MD = 210/⁠20 = 10.5.

Answer: (b) 10.65

Method. mean deviation about mean = Σf|x − x̅| / N using class marks

Σfx = 5 + 30 + 100 + 210 + 180 + 165 = 690, so x̅ = 690/⁠20 = 34.5.
|x − 34.5|: 29.5, 19.5, 9.5, 0.5, 10.5, 20.5.
f|x − 34.5|: 29.5, 39, 38, 3, 42, 61.5; total 213.
MD = 213/⁠20 = 10.65.

Answer: (d) Mode > Median > Mean

Method. order of mean, median, mode in a skewed distribution

Skewed to the left = negatively skewed: the long tail is on the left and pulls the mean lowest.
So Mean < Median < Mode, i.e. Mode > Median > Mean.

Answer: (c) 39

Method. grouped median = l + N/⁠2 − cff × h

N = 60, N/⁠2 = 30.
Cumulative frequencies: 4, 9, 20, 26, 31, … so the median class is 35–⁠40.
l = 35, cf = 26, f = 5, h = 5.
Median = 35 + 30 − 265 × 5 = 39.

Answer: (b) 27.73

Method. grouped mode = l + f₁ - f₀2f₁ - f₀ - f₂ × h

Highest frequency 11 ⇒ modal class 25–⁠30; f₀ = 5, f₁ = 11, f₂ = 6, h = 5.
Mode = 25 + 11 − 522 − 5 − 6 × 5 = 25 + 3011 = 27.73 (to two decimals).

Answer: (a) 1 only

Method. compute mean, median and mode in terms of x

Mean = 5x + 155 = x + 3.
Mode = x (occurs twice); median (3rd of 5 ordered values) = x + 2.
Mean > Mode (x + 3 > x): true. Median > Mean (x + 2 > x + 3): false.
Statement 1 only.

Answer: (c) 32 < g < 64

Method. GM of powers of 2 via exponents

Numbers are 2¹, 2², …, 2¹⁰.
g = (2^1+2+…+10)^1/⁠10 = 2^55/⁠10 = 2^5.5 = 32⁠√2⁠ ≈ 45.25.
So 32 < g < 64.

Answer: (c) 40

Method. HM of two numbers = 2ab/⁠(a + b)

2 × 60 × x60 + x = 48 ⇒ 120x = 2880 + 48x.
72x = 2880 ⇒ x = 40.

Answer: (c) Both 1 and 2

Method. definitions of AM and GM

AM = (x₁ + x₂ + … + x_N)/⁠N uses every value.
GM = (x₁x₂…x_N)^1/⁠N also uses every value.
So both 1 and 2 use all the data.

Answer: (a) 3

Method. median of a discrete frequency distribution via cumulative frequency

N = 198, so the median is the mean of the 99th and 100th values.
Cumulative frequencies: 4, 37, 113, 163, 189, 197, 198.
Values 38 to 113 all have 3 peas, so the 99th and 100th values are both 3.
Median = 3.

Answer: (d) m = 4n

Method. HM = 2mn/⁠(m + n), GM = ⁠√(mn)⁠

5 × 2mnm + n = 4⁠√(mn)⁠ ⇒ 10mn = 4(m + n)⁠√(mn)⁠ ⇒ 5⁠√(mn)⁠ = 2(m + n) (dividing by 2⁠√(mn)⁠).
Put t = ⁠√(m/⁠n)⁠ and divide by n: 5t = 2(t² + 1) ⇒ 2t² - 5t + 2 = 0 ⇒ (2t − 1)(t − 2) = 0 ⇒ t = 2 or 12.
t = 2 gives m = 4n (t = 1/⁠2 gives n = 4m, not offered).
Check m = 4, n = 1: HM = 8/⁠5, GM = 2, and 5(8/⁠5) = 8 = 4(2).

Answer: (b) 6, 18, 18, 18, 30

Method. compute mean, median and mode of each set

(a) mean = 72/⁠5 = 14.4, median = mode = 12.
(b) mean = (6 + 54 + 30)/⁠5 = 90/⁠5 = 18, median = 18, mode = 18.
(c) mean = 90/⁠5 = 18, median = 12, mode = 6.
(d) mean = 60/⁠5 = 12, median = mode = 6.
Only (b) has all three equal.

Answer: (c) 30000

Method. GM of a·x^k = a·G^k

GM of 3x_i⁴ = (Π 3x_i⁴)^1/⁠n = 3·(Πx_i)^4/⁠n = 3·G⁴.
= 3 × 10⁴ = 30000.

Answer: (d) 15

Method. mode = most frequent value

Without x, each of 10, 12, 13, 15 occurs twice, so there is no single mode.
For 15 to be the mode it must occur most often, so x = 15 (15 then occurs three times).

Answer: (b) 41, 43

Method. median of an even number of ordered observations is the mean of the two middle ones

There are 10 observations, so the median is the mean of the 5th and 6th: (x + 1) + (x + 3)2 = x + 2.
x + 2 = 42 gives x = 40.
5th = 41, 6th = 43 (consistent with the order 37 ≤ 41 ≤ 43 ≤ 46).

No correct option. This question is defective as printed in UPSC's paper.

Method. median of a discrete frequency distribution

N = 3 + 15 + 45 + 57 + 50 + 36 + 25 + 9 = 240.
Cumulative frequencies: 3, 18, 63, 120, 170, 206, 231, 240.
N is even, so the median is the mean of the 120th and 121st observations.
The 120th observation is 4 (cf reaches 120 at x = 4); the 121st is 5.
Median = 4 + 52 = 4.5, which is not among the options.

Defective as printed: N = 240, and the 120th and 121st observations are 4 and 5, so the median is 4.5, which no option prints. Two textbook shortcuts give different options: the value whose cumulative frequency first exceeds N/⁠2 gives 5, option (b); the value whose cumulative frequency is equal to or just greater than N/⁠2 gives 4, option (a). No option is defensible.

Answer: (a) 1 only

Method. sum of deviations from the mean; mean = median = mode for a symmetric distribution

1. Σ(x_i - x̅) = Σ x_i - nx̅ = nx̅ − nx̅ = 0. True.
2. For a symmetric distribution, mean = median = mode, so 'mean > median > mode' is false.
Only 1 is correct.

Answer: (a) PQ

Method. geometric mean of a ratio equals the ratio of the geometric means

P = (x₁ x₂ … x_n)^1/⁠n, Q = (y₁ y₂ … y_n)^1/⁠n.
GM of XY = (x₁y₁ · x₂y₂ … x_ny_n)^1/⁠n = (x₁ x₂ … x_n)^1/⁠n(y₁ y₂ … y_n)^1/⁠n = PQ.

Answer: (b) 4.9

Method. median of an even number of values

In ascending order: -4.8, 0, 2.3, 3.5, 3.9, 4.6, 5.2, 6.1, 7.6, 8.2, 9.3, 12.7.
There are 12 values, so the median is the mean of the 6th and 7th: 4.6 + 5.22 = 9.82 = 4.9.

Answer: (d) 40 km/⁠hr

Method. average speed = total distance ÷ total time

Time for first part = 530 = 16 h; time for second part = 1545 = 13 h.
Total time = 16 + 13 = 12 h; total distance = 20 km.
Average speed = 201/⁠2 = 40 km/⁠hr.

Answer: (b) 20%

Method. coefficient of variation = (σ/x̅) × 100%

CV = (8/⁠40) × 100% = 20%.

Answer: (d) 8

Method. mean deviation about the mean for grouped data, Σf|x − x̅|/⁠Σf

Total 100: 10 + 20 + f + 40 = 100 ⇒ f = 30.
Mid-points 5, 15, 25, 35; Σfx = 50 + 300 + 750 + 1400 = 2500; x̅ = 2500/⁠100 = 25.
|x − 25|: 20, 10, 0, 10; Σf|x − 25| = 200 + 200 + 0 + 400 = 800.
Mean deviation = 800/⁠100 = 8.

Answer: (d) 10

Method. σ² = Σf(x − x̅)²/⁠Σf for grouped data

From the previous item: f = 30, mid-points 5, 15, 25, 35, x̅ = 25.
(x − 25)²: 400, 100, 0, 100.
Σf(x − 25)² = 10(400) + 20(100) + 30(0) + 40(100) = 10000.
σ² = 10000/⁠100 = 100, σ = 10.

Answer: (b) II only

Method. units of measures of dispersion

I: variance is the mean of squared deviations, so its unit is the square of the data's unit (cm → cm²); the mean has the data's unit. I is false.
II: mean deviation and SD (= ⁠√variance)⁠ both have the data's own unit. II is true.

Answer: (b) Variance (X) < Variance (Y)

Method. Var = Σx²/⁠n − x̅²

X̅ = 200/⁠50 = 4, Var(X) = 900/⁠50 − 4² = 18 − 16 = 2.
Y̅ = 250/⁠50 = 5, Var(Y) = 1400/⁠50 − 5² = 28 − 25 = 3.
So Var(X) < Var(Y).

Answer: (a) Coefficient of variation of X is strictly more than coefficient of variation of Y.

Method. coefficient of variation = σ/⁠mean (× 100)

From the previous item: X̅ = 4, σ_X = ⁠√2;⁠ Y̅ = 5, σ_Y = ⁠√3.⁠
CV(X) = ⁠√2/⁠4⁠ ≈ 0.354, CV(Y) = ⁠√3/⁠5⁠ ≈ 0.346.
Compare squares: 2/⁠16 = 0.125 > 3/⁠25 = 0.12, so CV(X) > CV(Y) strictly.

Answer: (b) 2.5

Method. mean deviation about the mean of 1, 2, …, 10

Mean = (1 + 2 + … + 10)/⁠10 = 55/⁠10 = 5.5.
|x − 5.5| for x = 1, …, 10: 4.5, 3.5, 2.5, 1.5, 0.5, 0.5, 1.5, 2.5, 3.5, 4.5; sum = 2(12.5) = 25.
Mean deviation = 25/⁠10 = 2.5.

Answer: (b) 95

Method. σ² = (Σx_i²)/⁠n − M²

σ² = 19Σx_i² - M², so M² + σ² = 19Σx_i².
M² + σ² = 855/⁠9 = 95.

Answer: (c) 6

Method. mean deviation about the mean of five ungrouped values

Largest five marks: 50, 47, 41, 35, 32.
Mean = (50 + 47 + 41 + 35 + 32)/⁠5 = 205/⁠5 = 41.
|x − 41|: 9, 6, 0, 6, 9; sum = 30.
Mean deviation = 30/⁠5 = 6.

Answer: (d) 46.8

Method. variance as the mean of squared deviations from the mean

From the previous item, the five values are 50, 47, 41, 35, 32 with mean 41.
(x − 41)²: 81, 36, 0, 36, 81; sum = 234.
Variance = 234/⁠5 = 46.8.

Answer: (c) 112

Method. Σ(ax_i + b) = aΣx_i + nb

Σx_i = 10 × 20 = 200.
Σ3x_i - 45 = 3 × 200 − 4 × 105 = 5605 = 112.

Answer: (d) 4

Method. σ² = Σx²/⁠n − x̅²

σ² = 1616010 − 40² = 1616 − 1600 = 16.
σ = 4.

Answer: (c) 10 and 13

Method. use Σx = nx̅ and Σx² = n(σ² + ⁠x̅⁠²)

Σx = 5 × 14 = 70; the other two sum to 70 − (11 + 16 + 20) = 23.
Σx² = 5(13.2 + 196) = 1046; the other two have squares summing to 1046 − (121 + 256 + 400) = 269.
All options sum to 23; check squares: 10² + 13² = 100 + 169 = 269 (8, 15 gives 289; 9, 14 gives 277; 11, 12 gives 265).
The other two observations are 10 and 13.

Answer: (d) Biology

Method. compare variability by coefficient of variation CV = (SD/⁠mean) × 100

Mathematics: 15/⁠40 × 100 = 37.5%.
Physics: 12/⁠28 × 100 ≈ 42.9%.
Chemistry: 14/⁠38 × 100 ≈ 36.8%.
Biology: 16/⁠36 × 100 ≈ 44.4%.
The largest CV is Biology (it also has the largest SD).

Answer: (a) 37.5%

Method. CV = (SD/⁠mean) × 100

CV (Mathematics) = 1540 × 100 = 37.5%.

Answer: (c) 7

Method. variance = Σx²n − x̅²

Let the fifth value be x. Σx = 13 + x, Σx² = 49 + x².
49 + x²5 − (13 + x)²25 = 3.6 ⇒ 5(49 + x²) − (13 + x)² = 90.
4x² - 26x − 14 = 0 ⇒ 2x² - 13x − 7 = 0 ⇒ (x − 7)(2x + 1) = 0.
x is positive, so x = 7.

Answer: (b) 0.22

Method. coefficient of MD = mean deviation ÷ mean

Sum = 300, mean = 30.
|deviations|: 9, 4, 7, 9, 4, 7, 10, 10, 3, 3; sum = 66.
MD = 66/⁠10 = 6.6; coefficient = 6.6/⁠30 = 0.22.

Answer: (c) Standard deviation

Method. standard deviation is the measure of dispersion built on the mean

SD = ⁠√(Σ(x⁠ −⁠ ⁠x̅⁠)²/⁠n) uses every value and is measured about the mean, squaring deviations instead of ignoring signs.
Range uses two values and quartile deviation the quartiles; mean deviation ignores signs.
So with the mean available, SD is the most reliable measure.

Answer: (a) 5

Method. mean deviation about the mean

Numbers 2, 4, …, 20; mean = 2 + 202 = 11.
|deviations|: 9, 7, 5, 3, 1, 1, 3, 5, 7, 9; sum = 50.
Mean deviation = 50/⁠10 = 5.

Answer: (b) 33

Method. variance = Σx²n − x̅²

x̅ = 110/⁠10 = 11.
Variance = 154010 − 11² = 154 − 121 = 33.

Answer: (a) 2025

Method. CV = (σ/x̅) × 100

45 = σ100 × 100 ⇒ σ = 45.
Variance = σ² = 45² = 2025.

Answer: (b) 2

Method. symmetric data about 0 have mean 0, so σ² = Σx²n

The observations pair off as ±, so the mean is 0.
Σx² = 2(6 + 5 + 4 + 1) = 32, n = 8.
σ² = 328 = 4, so σ = 2.

Answer: (d) 200

Method. coefficient of variation = σx̅ × 100

x̅ = 2010 = 2.
σ² = 20010 − 2² = 16, so σ = 4.
CV = 42 × 100 = 200.

Answer: (a) 20

Method. variance from deviations about an arbitrary point: σ² = Σ(x − A)²n − (x̅ − A)²

With A = 50: Σ(x − 50)²n = 500010 = 500.
x̅ − A = 60 − 50 = 10.
σ² = 500 − 10² = 400, so σ = 20.

Answer: (b) 9, 4

Method. use the mean for the sum and the variance for the sum of squares of the missing pair

Sum = 5 × 4.4 = 22, so a + b = 22 − (1 + 2 + 6) = 13.
Σx²5 = σ² + x̅² = 8.24 + 19.36 = 27.6, so Σx² = 138 and a² + b² = 138 − (1 + 4 + 36) = 97.
ab = (a + b)² - (a² + b²)2 = 169 − 972 = 36.
a, b are roots of t² - 13t + 36 = 0: t = 9, 4.

Shortcut. Check options: only 9 + 4 = 13 matches the required sum.

Answer: (c) 2483

Method. V + M² = Σx²n

V = Σx²n − M², so V + M² = Σx²n.
Σx² for x = 1 to 15 is 15 × 16 × 316 = 1240.
V + M² = 124015 = 2483.

Answer: (c) 299.25

Method. variance of first n natural numbers is n² - 112; multiplying by k multiplies variance by k²

Taking the numbers 1, 2, …, 20 (n = 20): Var = 20² - 112 = 39912 = 33.25.
Multiplying each by 3 multiplies the variance by 3² = 9.
New variance = 9 × 33.25 = 299.25.
(Excluding 1 and 20 would give 242.25, which is not an option; the inclusive reading is intended.)

Answer: (d) None of the above

Method. combined mean and combined variance of two groups

Combined mean = 500 × 1860 + 600 × 17501100 = 19800001100 = ₹ 1800.
d_A = 1860 − 1800 = 60, d_B = 1750 − 1800 = -50.
Combined variance = 500(81 + 60²) + 600(100 + 50²)1100 = 1840500 + 15600001100 = 34005001100 ≈ 3091.4.
The mean ₹ 1800 matches option (c), but the variance is about 3091, not 81; no option (a)–⁠(c) fits, so the answer is 'None of the above'.

Answer: (a) y ≥ x

Method. standard deviation is never less than mean deviation about the mean

Let d_i = |x_i - x̅|. Then y² - x² = 1n Σ d_i² - (1n Σ d_i)² = 1n Σ (d_i - x)² ≥ 0.
So y ≥ x, with equality when all |x_i - x̅| are equal.
Equality can occur with unequal values: for 1, 3, mean deviation = 1 and SD = 1. So x < y (option d) is not always true; y ≥ x is.

Answer: (b) σ = ⁠√(N²⁠ - 112)

Method. σ² = Σx²N − x̅² for 1, 2, …, N

x̅ = N + 12, Σx²N = (N + 1)(2N + 1)6.
σ² = (N + 1)(2N + 1)6 − (N + 1)²4 = (N + 1)[2(2N + 1) − 3(N + 1)]12 = (N + 1)(N − 1)12 = N² - 112.
σ = ⁠√(N²⁠ - 112).

Answer: (b) 2 and 3 only

Method. properties of measures of dispersion

(1) CV = σx̅ × 100 is a ratio of two quantities in the same unit, so it is unit-free: false.
(2) Range = largest − smallest is a measure of dispersion: true.
(3) Σ|x_i - A| is least when A is the median: true.
Correct: 2 and 3 only.

Answer: (d) 24

Method. zero standard deviation means all observations are equal

σ = 0 means Σ(x_i - x̅)² = 0, so every x_i = x̅ = 24.
Any five of them, including the smallest five, have mean 24.

Answer: (a) Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1

Method. range uses only the two extreme values

Range = largest value − smallest value, so it depends only on the two extremes.
A single extreme value changes it a lot (Statement 2 is true), and it ignores every other observation.
That sensitivity is exactly why range is a poor measure of dispersion (Statement 1 is true and is explained by Statement 2).

Answer: (c) 4 × Standard deviation = 5 × Mean deviation

Method. empirical relation MD ≈ 45 SD for moderately asymmetric data

For moderately non-symmetrical data: Mean deviation ≈ 45 × Standard deviation.
Multiplying by 5: 5 × MD = 4 × SD.
This is option (c), 4 × Standard deviation = 5 × Mean deviation.

Answer: (b) 100

Method. Var(ax + b) = a²Var(x)

New observation y = 2x − 7.
Subtracting 7 leaves the variance unchanged; multiplying by 2 multiplies it by 2² = 4.
Var(y) = 4 × 25 = 100.

Answer: (d) 13656

Method. linearity of summation, Σ(ax_i + b) = aΣx_i + nb

Σx_i = 20 × 1.414 = 28.28.
Σ100(2x_i + 4) = 100(2Σx_i + 4 × 20) = 100(56.56 + 80) = 100 × 136.56 = 13656.

Answer: (d) 50

Method. SD scales by |k| when every observation is multiplied by k

Multiplying each observation by 5 multiplies every deviation from the mean by 5.
New SD = 5 × 10 = 50.

Answer: (a) 10

Method. SD is unchanged when a constant is added to every observation

Adding 20 shifts every observation and the mean by 20, so every deviation x − x̅ is unchanged.
New SD = 10.

Answer: (a) 2.25

Method. mean under the change y = (x − a)/⁠h

New observation y = (x − 5)/⁠20, so y̅ = (x̅ − 5)/⁠20.
y̅ = (50 − 5)/⁠20 = 45/⁠20 = 2.25.

Answer: (b) 0.5

Method. SD is unchanged by adding a constant and divided by |h| when dividing by h

New observation y = (x + 5)/⁠20.
Adding 5 does not change the SD; dividing by 20 divides it by 20.
σ_y = 10/⁠20 = 0.5.

Answer: (b) 22

Method. mean under a linear change y = a + bx

New observation y_i = 44 − 4x_i.
New mean = 44 − 4 × 5.5 = 44 − 22 = 22.

Answer: (b) M + k

Method. mean shifts with a change of origin

Mean of (x_i - k) = x̅ − k = M.
So x̅ = M + k.

Answer: (c) 55, 10

Method. adding a constant shifts the mean but not the standard deviation

New mean = 50 + 5 = 55.
Every deviation from the mean is unchanged, so the SD stays 10.

Answer: (c) 75

Method. range under a linear transformation Y = a + bX is |b| × range of X

Max Y − min Y = (40 + 3 X_max) − (40 + 3 X_min) = 3(X_max - X_min).
Range of Y = 3 × 25 = 75.

Answer: (c) 11.25, 2.5

Method. effect of shifting and scaling on mean and standard deviation

New observation y = x − 54.
New mean = 50 − 54 = 454 = 11.25.
Shifting does not change SD; dividing by 4 divides it by 4: new SD = 104 = 2.5.

Answer: (c) 1 and 3 only

Method. effect of change of origin and scale on mean and SD

(1) Mean of x + 10 = x̅ + 10: true.
(2) SD of x + 10 = SD of x (change of origin does not affect SD): false.
(3) Mean of 2x = 2x̅: true.
Correct: 1 and 3 only.

Answer: (b) 4

Method. variance is unchanged when a constant is added to every observation

Adding 2 to each x_i raises the mean by 2, so every deviation x_i - x̅ is unchanged.
Variance = 1n Σ (x_i - x̅)² is therefore unchanged.
New variance = 4.

Answer: (d) x̅ − ac

Method. mean is linear: mean of a + cy is a + cy̅

x̅ = 1nΣ(a + cy_i) = a + cy̅.
So y̅ = x̅ − ac.

Answer: (d) Neither 1 nor 2

Method. effect of change of origin and scale on mean and variance

(1) Mean of a + cx is a + cx̅: the mean changes with both origin and scale. False.
(2) Var(a + cx) = c²Var(x): variance is independent of origin but not of scale, the reverse of what is stated. False.
Neither is correct.

Answer: (b) 20

Method. adding a constant shifts the mean but not the SD

New mean = 5 + 5 = 10; new SD = 2 (unchanged by change of origin).
CV = σx̅ × 100 = 210 × 100 = 20.

Answer: (b) 2 only

Method. effect of origin and scale on variance; CV is unit-free

(1) Var(x + a) = Var(x) but Var(cx) = c²Var(x): variance is unaffected by change of origin but affected by change of scale, so (1) is false.
(2) Reading 'coefficient of variance' as the coefficient of variation, CV = σx̅ × 100 is a ratio of two quantities in the same unit, hence independent of the unit: true.
Only 2 is correct.

Answer: (d) 45

Method. variance scales by the square of a multiplying constant

If every x_i becomes 3x_i, then Var(3x) = 3² Var(x).
New variance = 9 × 5 = 45.

Answer: (a) -r

Method. r(ax + b, cy + d) = r × sign(ac)

Correlation is unchanged in size by change of origin and scale; only the sign of the product of the scale factors matters.
Here a = 3, c = -3, ac = -9 < 0.
Required correlation = -r.

Answer: (c) 0.7

Method. correlation is unaffected by a change of origin

After 25 years each husband's age is x + 25 and each wife's age is y + 25.
Adding a constant to every value of a variable does not change deviations from the mean, so r is unchanged.
r = 0.7.

Answer: (c) both change of scale and change of origin

Method. correlation coefficient is unaffected by change of origin and scale

If u = (x − a)/⁠h and v = (y − b)/⁠k with h, k > 0, then r(u, v) = r(x, y).
Shifting (origin) cancels in the deviations, and scaling cancels between covariance and the two SDs.
So r is independent of both change of origin and change of scale.

Answer: (c) 0

Method. independent variables have zero covariance

If X and Y are independent, E(XY) = E(X)E(Y), so Cov(X, Y) = E(XY) − E(X)E(Y) = 0.
r = Cov(X, Y)σ_X σ_Y = 0.

Answer: (d) 0.6

Method. correlation is unchanged by change of origin and positive change of scale

Z = X + 5 shifts the origin only; W = 13Y rescales by a positive factor.
r(Z, W) = r(X, Y) × 1 × 13|1| · |13| = r(X, Y) = 0.6.

Answer: (d) 36%

Method. unexplained variation = 1 − r²

Proportion explained by the linear regression = r² = 0.8² = 0.64.
Proportion not explained = 1 − 0.64 = 0.36, i.e. 36%.

Answer: (d) -1 and 1 only

Method. points on a line of non-zero slope give perfect correlation

All points satisfy y = mx + c with m ≠ 0, so y_i - y̅ = m(x_i - x̅) for every i.
r = m Σ(x_i - x̅)²⁠√(Σ(x_i⁠ - ⁠x̅⁠)²) · |m| ⁠√(Σ(x_i⁠ - ⁠x̅⁠)²) = m|m|.
So r = 1 if m > 0 and r = -1 if m < 0; r = 0 is impossible. Possible values: -1 and 1 only.

Answer: (c) Insurance companies' profits and the number of claims they have to pay

Method. negative correlation: one variable rises as the other falls

(a) Ages of husbands and wives rise together: positive.
(b) Shoe size and intelligence are unrelated: about zero.
(c) More claims paid means lower profit: negative.
(d) More rainfall (within normal limits) gives higher yield: positive.
Negative correlation: (c).

Answer: (d) 81

Method. Cov(x, y) = r σ_x σ_y

σ_y = ⁠√25⁠ = 5.
27 = 0.6 × σ_x × 5 = 3σ_x, so σ_x = 9.
Variance of x = 9² = 81.

Answer: (b) X = ⁠X̅⁠ and Y = ⁠Y̅⁠

Method. when r = 0, b_yx = b_xy = 0

b_yx = rσ_y/⁠σ_x = 0 and b_xy = rσ_x/⁠σ_y = 0.
y on x: Y − Y̅ = 0·(X − X̅) ⇒ Y = ⁠Y̅⁠.
x on y: X − X̅ = 0·(Y − Y̅) ⇒ X = ⁠X̅⁠.
The lines are X = ⁠X̅⁠ and Y = ⁠Y̅⁠ (perpendicular). Check: (c) gives r² = 0.4 × 1.6 = 0.64 ≠ 0, and (a) gives r² = 2/⁠3 ≠ 0.

Answer: (d) 1

Method. b_yx = r σ_y/⁠σ_x

b_yx = 0.8 × 5/⁠4 = 1.

Answer: (a) -⁠√(23)⁠

Method. identify the regressions by b_yxb_xy ≤ 1, then r = ±⁠√(b_yxb_xy)⁠ with the sign of the coefficients

Try x + y + 11 = 0 as y on x: b_yx = -1; then 2x + 3y + 4 = 0 as x on y: b_xy = -3/⁠2. Product 3/⁠2 > 1, impossible.
So 2x + 3y + 4 = 0 is y on x: y = (-2x − 4)/⁠3, b_yx = -2/⁠3.
And x + y + 11 = 0 is x on y: x = -y − 11, b_xy = -1.
r² = (-2/⁠3)(-1) = 2/⁠3; both coefficients negative, so r = -⁠√(23)⁠.

Answer: (d) 5

Method. choosing which line is y on x from the condition b_yxb_xy ≤ 1

Try x − 3y + 4 = 0 as y on x: b_yx = 1/⁠3; then 2x − 7y + 8 = 0 as x on y gives x = (7y − 8)/⁠2, b_xy = 7/⁠2. Product = 7/⁠6 > 1, impossible.
So 2x − 7y + 8 = 0 is y on x: y = (2x + 8)/⁠7, b_yx = 2/⁠7.
And x − 3y + 4 = 0 is x on y: x = 3y − 4, b_xy = 3. Product = 6/⁠7 ≤ 1 (holds).
b_xy + 7b_yx = 3 + 2 = 5.

Answer: (b) 11

Method. b = Σxy − nx̅y̅Σx² − nx̅², a = y̅ − bx̅

Σx = -4 − 1 + 2 + 3 = 0, so x̅ = 0; Σy = 11, so y̅ = 11/⁠4.
Σxy = -4 − 2 + 14 + 3 = 11; Σx² = 16 + 1 + 4 + 9 = 30.
b = 11 − 030 − 0 = 1130; a = 114 − 0 = 114.
2a + 15b = 112 + 112 = 11.

Answer: (a) 191

Method. angle between two lines, then tan3θ = 3tanθ − tan³θ1 − 3tan²θ

Slopes: m₁ = -1/⁠2, m₂ = -2/⁠3.
tanθ = |m₁ - m₂1 + m₁m₂| = |-1/⁠2 + 2/⁠31 + 1/⁠3| = 1/⁠64/⁠3 = 18.
tan3θ = 3/⁠8 − 1/⁠5121 − 3/⁠64 = 191/⁠51261/⁠64 = 191488.
488tan3θ = 191.

Answer: (c) 5

Method. identify the regression lines using b_yx·b_xy ≤ 1, then use x on y

Try x + 4y + 1 = 0 as y on x: b_yx = -1/⁠4; and 4x + 9y + 7 = 0 as x on y: x = -94y − 74, b_xy = -9/⁠4.
Product = 9/⁠16 ≤ 1, valid. (The other assignment gives (-4)(-4/⁠9) = 16/⁠9 > 1, impossible.)
To estimate x from y use x on y: x = -94(-3) − 74 = 27 − 74 = 5.

Answer: (a) x − 4y + 5 = 0

Method. with only two points the least-squares line passes through both

Two points fit a straight line exactly, so the regression line of y on x is the line through them.
Slope = (2 − 1)/⁠(3 − (-1)) = 1/⁠4.
y − 1 = 14(x + 1) ⇒ 4y − 4 = x + 1 ⇒ x − 4y + 5 = 0.
Check (3, 2): 3 − 8 + 5 = 0.

Answer: (c) Both 1 and 2

Method. identify regression lines by b_yx·b_xy ≤ 1

Line 1: 3x − 4y + 8 = 0 ⇒ y = 34x + 2; Line 2: 4x − 3y − 1 = 0 ⇒ x = 34y + 14.
If line 1 is y on x and line 2 is x on y: b_yxb_xy = 34 × 34 = 916 ≤ 1, valid.
The other assignment gives b_yx = b_xy = 43, product 169 > 1, impossible.
So both statements are correct.

Answer: (a) 1 only

Method. r = ±⁠√(b_yxb_xy)⁠; means lie at the intersection of the lines

b_yx = 34, b_xy = 34 (from the previous item) ⇒ r² = 916, r = 34 (both slopes positive).
Means: solve 3x − 4y = -8 and 4x − 3y = 1 ⇒ y = 5, x = 4.
So x̅ = 4, y̅ = 5, not 3 and 4: statement 1 only.

Answer: (b) Y = 1.12X − 5.8

Method. regression line of Y on X: Y − Y̅ = r σ_Yσ_X (X − X̅)

b_YX = 0.8 × 3.52.5 = 1.12.
Y − 67 = 1.12(X − 65) = 1.12X − 72.8.
Y = 1.12X − 5.8.

Answer: (c) 359, 203, -0.5

Method. regression lines meet at (x̅, y̅); choose the assignment with b_yx b_xy ≤ 1

Solve 6x + y = 30 and 3x + 2y = 25: y = 30 − 6x, so 3x + 60 − 12x = 25, x = 359; y = 30 − 703 = 203.
So x̅ = 359, y̅ = 203.
If 6x + y = 30 were Y on X: b_yx = -6, and 3x + 2y = 25 as X on Y gives b_xy = -23; product 4 > 1, impossible.
So Y on X is 3x + 2y = 25: b_yx = -32; X on Y is 6x + y = 30: b_xy = -16.
r² = 32 × 16 = 14; both coefficients negative, so r = -0.5.

Answer: (b) -124

Method. product of the regression coefficients equals r²

b_yx · b_xy = r² = (-12)² = 14.
b_xy = 1/⁠4-6 = -124.
Both coefficients carry the sign of r (negative), as required.

Answer: (d) Neither 1 nor 2

Method. regression lines when r = 0 and when r = ±1

(1) r = 0 gives b_yx = b_xy = 0: the lines are y = y̅ (horizontal) and x = x̅ (vertical), which are perpendicular, not parallel. False.
(2) r = +1 gives b_yx = σ_yσ_x and b_xy = σ_xσ_y: both lines become y − y̅ = σ_yσ_x(x − x̅), i.e. they coincide, not perpendicular. False.
Neither is correct.

Answer: (c) 13 and 17

Method. both regression lines pass through (x̅, y̅)

Solve 4x − 5y = -33 and 20x − 9y = 107 together.
First × 5: 20x − 25y = -165. Subtract from the second: 16y = 272, so y = 17.
4x = 5(17) − 33 = 52, so x = 13.
x̅ = 13, y̅ = 17.

Answer: (a) (1, 1)

Method. b_yx and b_xy have the same sign and b_yxb_xy = r² ≤ 1

(a) (1, 1): same sign, product 1, r = 1 (perfect positive correlation): admissible.
(b) (-1, 1) and (c) (-12, 2): opposite signs, impossible since both coefficients have the sign of r.
(d) (13, 103): product 109 > 1, impossible.
Only (a) can occur.