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Saturday, 31 October

JEE Main 2024 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 1 February 2024, Shift 2, Maths Q19

Question 19 of 90 in this shift, Maths question 19 of 30, Section A.

Consider 1010 observations x1,x2,…,x10x_1, x_2, \ldots, x_{10} such that ∑i=110(xi−α)=2\sum_{i=1}^{10}(x_i-\alpha)=2 and ∑i=110(xi−β)2=40\sum_{i=1}^{10}(x_i-\beta)^2=40, where α\alpha, β\beta are positive integers. Let the mean and the variance of the observations be 65\frac{6}{5} and 8425\frac{84}{25} respectively. Then βα\frac{\beta}{\alpha} is equal to :
  1. (1)11
  2. (2)22Official answer
  3. (3)32\frac{3}{2}
  4. (4)52\frac{5}{2}

Official answer

Option 2

NTA final key.

Chapter
Statistics
Topic
Variance and SD
Idea tested
Variance via Arbitrary Point Deviations

Same idea in other shifts

Asked 3× in all
  1. 3 Aug 2021, Shift 1 · Q77If ∑i=116ai−6=8\sum_{i=1}^{16} a_i - 6 = 8 and ∑i=116ai−6 2=40\sum_{i=1}^{16} a_i - 6^{\,2} = 40, then the standard deviation of a1,a2,…,a16a_1, a_2, \ldots, a_{16} isEasySingle correct
  2. 3 Sep 2020, Shift 2 · Q67Let xix_i (1≤i≤10)(1 \leq i \leq 10) be ten observations of a random variable X. If ∑i=110(xi−p)=3\sum_{i=1}^{10}(x_i - p) = 3 and…EasySingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.