Shifting, scaling and correcting data

Adding the same number a to every observation (a positive or negative) shifts the mean by a but leaves every deviation, and so the variance and standard deviation, unchanged.

Multiplying every observation by a non-zero k multiplies the mean by k, the variance by k² and the standard deviation by |k|.

Trial openers (made up): a bonus of 5 runs on every innings makes Arjun's mean 55 with σ still ≈ 4.82. Counting 2 points per run makes the mean 100, the variance 4 × 23.25 = 93 and σ ≈ 9.64.

NCERT's worked results: variance 5 of 20 observations becomes 2² × 5 = 20 when each is doubled.

Correcting a wrong entry: rebuild Σx from n x̄ and Σx² from n(σ² + x̄²), swap the wrong value for the right one in both, then recompute. Made-up case: 20 fielding scores with mean 50 and σ = 5 had a 60 entered as 40. Σx = 1000 → 1020 and Σx² = 50500 → 52500, so the correct mean is 51 and σ² = 2625 − 51² = 24, σ ≈ 4.90.

NCERT's worked results: 100 observations with mean 40 and σ = 5.1, one of them 40 misread as 50, correct to mean 39.9 and σ = 5.

Finding missing values: the mean fixes their sum and the variance fixes the sum of their squares. Arjun's 8 innings have mean 50 and variance 23.25; if six are 42, 48, 50, 52, 55 and 45, the other two satisfy x + y = 108 and x² + y² = 5864, so (x − y)² = 64 and they are 58 and 50. NCERT's own example of this kind gives the missing observations 4 and 9.