Shift the scores, stretch the scores
"Shift the scores, stretch the scores" is a free, interactive Class 11 Maths simulation for JEE, built on the lesson "Shifting, scaling and correcting data" in Statistics. The dots are Arjun's 8 innings (made-up data), turned into k × runs + c. The bars below are each innings' deviation from the new mean, on a fixed scale. Try c first, then k. For y = kx + c: the mean becomes k x̄ + c, the variance k² σ² and the standard deviation |k| σ. Adding c moves the data without spreading it. JEE Main asked 32 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
From the lesson Shifting, scaling and correcting data in Statistics. Change the values and watch what happens.
In JEE Main
42 shifts analysed
32 questions from Statistics across the 42 JEE Main shifts Lumi has analysed (2017–2026).
The idea behind it
NCERT § "Miscellaneous Examples"
- 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.
Questions
3 answers
- What does the "Shift the scores, stretch the scores" simulation show?
- The dots are Arjun's 8 innings (made-up data), turned into k × runs + c. The bars below are each innings' deviation from the new mean, on a fixed scale. Try c first, then k.
- Does JEE Main ask questions from Statistics?
- Yes. JEE Main asked 32 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
- Is this simulation free?
- Yes. It is free and runs in the browser on a phone or a laptop. It belongs to the lesson "Shifting, scaling and correcting data" in Statistics (Class 11 Maths).
Embed this simulation
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