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MathsEasyClass 11

Statistics — Variance and Standard Deviation

Statistics

35

JEE Qs

5%

Hard

60

min

Master the shortcut computational formulas and understand the effect of transformations (change of origin and scale) on variance and standard deviation to solve problems quickly and accurately.

🧮 Key Formulas

Variance (ungrouped data), σ^2 = (1/n) * Σ(x_i - x̄)^2
Variance (ungrouped data, computational), σ^2 = (1/n) * Σ(x_i^2) - (x̄)^2
Standard Deviation (ungrouped data), σ = sqrt((1/n) * Σ(x_i - x̄)^2)
Variance (grouped data), σ^2 = (1/N) * Σ(f_i * (x_i - x̄)^2), where N = Σf_i
Variance (grouped data, computational), σ^2 = (1/N) * Σ(f_i * x_i^2) - (x̄)^2
Standard Deviation (grouped data), σ = sqrt((1/N) * Σ(f_i * (x_i - x̄)^2))
Coefficient of Variation (CV) = (σ / x̄) * 100%

✅ Key Points for JEE

  • 1Variance (σ^2) and Standard Deviation (σ) are measures of dispersion, indicating how spread out data points are from the mean; higher values mean greater spread.
  • 2Standard Deviation is the positive square root of variance, making it interpretable in the original units of the data.
  • 3Variance and Standard Deviation are unaffected by a change of origin (adding/subtracting a constant to each observation).
  • 4If each observation is multiplied by a constant 'b', the new variance becomes b^2 * old variance, and the new standard deviation becomes |b| * old standard deviation.
  • 5The computational formulas (using Σx_i^2 or Σf_i*x_i^2) are generally faster and less prone to rounding errors in exam calculations compared to direct deviation methods.

⚠️ Common Mistakes

  • Calculation errors, especially with negative numbers, squaring, or summing large quantities.
  • Confusing variance with standard deviation, or forgetting to take the square root for standard deviation.
  • Incorrectly applying the properties of change of origin and scale to variance/standard deviation.
  • Using formulas for ungrouped data when dealing with grouped frequency distribution or vice-versa.

📝 Practice Questions

See all

Q1. Let x1, x2, … , x10 be ten observations such that ∑10i=1 (xi −2) = 30, ∑10i=1 (xi −β)2 = 98, β > 2, and their variance is 4 . If μ and σ2 are respectively the mean and the variance of 2 (x1 −1) + 4β , 5 2 (x2 −1) + 4β, … . , 2 (x10 −1) + 4β , then βμσ2 is equal to : (1) 100 (2) 120 (3) 110 (4) 90

2025·MCQMedium

Q11.The area of the region {(x, y) : x2 + 4x + 2 ≤y ≤|x + 2|} is equal to (1) 7 (2) 5 (3) 24/5 (4) 20/3

2025·MCQMedium

Q5. Marks obtains by all the students of class 12 are presented in a freqency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12 . If the number of students whose marks are less than 12 is 18 , then the total number of students is (1) 52 (2) 48 (3) 44 (4) 40

2025·MCQMedium

Q86. X α 1 0 −3 Let the mean and the standard deviation of the probability distribution be μ and σ, P(X) 31 K 16 41 respectively. If σ −μ = 2, then σ + μ is equal to________ JEE Main 2024 (05 Apr Shift 2) JEE Main Previous Year Paper

2024·NumericalMedium

Q69.If the mean of the following probability distribution of a random variable X : X 0 2 4 6 8 46 is , then the variance of the distribution is P(X) a 2a a + b 2b 3b 9 (1) 173 (2) 566 27 81 (3) 151 (4) 581 27 81

2024·MCQMedium

Q69.Consider 10 observation 𝑥1, 𝑥2, . .. 𝑥10, such that ∑𝑖=10 1 𝑥𝑖−𝛼= 2 and ∑𝑖=10 1 𝑥𝑖−𝛽2 = 40, where 𝛼, 𝛽 are 6 84 𝛽 positive integers. Let the mean and the variance of the observations be and respectively. The is equal to: 5 25 𝛼 (1) 2 (2) 3 2 (3) 5 (4) 1 2

2024·MCQMedium

NCERT Chapters

  • Class 11 Mathematics Ch 15: Statistics