Miscellaneous Exercise answers: Statistics

Open NCERT p. 286Matches NCERT’s answer

1. Let the remaining two observations be a and b.
2. The 8 observations add to 8 × 9 = 72. The six known ones add to 60, so a + b = 12.
3. The sum of squares of all 8 observations is 8 × (9.25 + 9²) = 722. The six known squares add to 642, so a² + b² = 80.
4. Using (a+b)² = a² + b² + 2ab: 144 = 80 + 2ab, so ab = 32.
5. a and b are the roots of t² − 12t + 32 = 0, which factorises as (t−4)(t−8) = 0.
6. So the two remaining observations are 4 and 8.

AnswerThe remaining two observations are 4 and 8.

Watch this explained “What survives”, 10:01 into Carrying the frequencies through, for discrete and for grouped data

Open NCERT p. 286Matches NCERT’s answer

1. Let the remaining two observations be a and b.
2. The 7 observations add to 7 × 8 = 56. The five known ones add to 42, so a + b = 14.
3. The sum of squares of all 7 observations is 7 × (16 + 8²) = 560. The five known squares add to 460, so a² + b² = 100.
4. Using (a+b)² = a² + b² + 2ab: 196 = 100 + 2ab, so ab = 48.
5. a and b are the roots of t² − 14t + 48 = 0, which factorises as (t−6)(t−8) = 0.
6. So the two remaining observations are 6 and 8.

AnswerThe remaining two observations are 6 and 8.

Watch this explained “What survives”, 10:01 into Carrying the frequencies through, for discrete and for grouped data

Open NCERT p. 286Matches NCERT’s answer

1. When every observation is multiplied by 3, the mean is also multiplied by 3.
2. New mean = 8 × 3 = 24.
3. The standard deviation is also multiplied by 3 (never by 3², since a standard deviation is a distance, not a squared distance).
4. New standard deviation = 4 × 3 = 12.

AnswerNew mean = 24, new standard deviation = 12.

Watch this explained “The scale, standing on its own”, 15:30 into Shifting and scaling the observations to make the arithmetic small

Open NCERT p. 286One way to think about it

1. The mean of ax₁, ax₂, …, axₙ is (ax₁+ax₂+…+axₙ)/n = a(x₁+x₂+…+xₙ)/n = a × x̄.
2. The deviation of each new observation axᵢ from the new mean a x̄ is axᵢ − a x̄ = a(xᵢ − x̄).
3. Squaring this deviation gives a²(xᵢ − x̄)².
4. Average these squared deviations over all n observations: the new variance is a² × [(1/n)Σ(xᵢ − x̄)²] = a²σ², since the bracketed part is exactly σ², the variance of the original observations.
5. So the mean of ax₁,…,axₙ is a x̄ and their variance is a²σ².

In shortThe mean of the scaled observations is a x̄ and their variance is a²σ².

Watch this explained “The square gets the square”, 3:52 into Shifting and scaling the observations to make the arithmetic small

Open NCERT p. 286Checked by computer

(i) If wrong item is omitted

1. The total of the 20 observations is 20 × 10 = 200, and the total of their squares is 20 × (2² + 10²) = 2080.
2. Removing the wrong item 8 leaves 19 observations totalling 200 − 8 = 192, so the correct mean is 192 ÷ 19.
3. The total of squares without it is 2080 − 8² = 2016.
4. Correct variance = 2016/19 − (192/19)² = 1440/361.
5. Correct standard deviation = √(1440/361) = 12√10⁄19 ≈ 2.00.

Answercorrect mean = 192/19 ≈ 10.11, correct standard deviation = 12√10⁄19 ≈ 2.00

(ii) If it is replaced by 12

1. Replace the wrong 8 by 12: the total becomes 200 − 8 + 12 = 204, still over 20 observations, so the correct mean is 204 ÷ 20 = 10.2.
2. The total of squares becomes 2080 − 8² + 12² = 2160.
3. Correct variance = 2160/20 − (10.2)² = 108 − 104.04 = 3.96.
4. Correct standard deviation = √3.96 = 3√11⁄5 ≈ 1.99.
5. The answer key at the back of the book prints 1.99 and 1.98; the exact values are √3.989… = 1.997… and √3.96 = 1.989…, which round to 2.00 and 1.99, so the key cut the decimals short.

Answercorrect mean = 51/5 = 10.2, correct standard deviation = 3√11⁄5 ≈ 1.99

Watch this explained “What survives”, 10:01 into Carrying the frequencies through, for discrete and for grouped data

Open NCERT p. 286Matches NCERT’s answer

1. The total of all 100 observations is 100 × 20 = 2000, and the total of their squares is 100 × (3² + 20²) = 40900.
2. Remove the three incorrect readings 21, 21, 18: their sum is 60, so the remaining 97 observations total 2000 − 60 = 1940.
3. Mean = 1940 ÷ 97 = 20.
4. The sum of the squares of the three incorrect readings is 21² + 21² + 18² = 1206, so the remaining sum of squares is 40900 − 1206 = 39694.
5. Variance = 39694/97 − 20² = 894/97.
6. Standard deviation = √(894/97) ≈ 3.04.

AnswerMean = 20, standard deviation = √(894/97) ≈ 3.04.

Watch this explained “What survives”, 10:01 into Carrying the frequencies through, for discrete and for grouped data