NTA Abhyas JEE Main2020MathematicsStatisticsPractice

Mean of the numbers 1,2 , 3 , … … … … , 9 with respective weights 1 2 + 1 , 2 2 + 2 , 3 2 + 3 , … , 9 2 + 9 is

Correct answer

Step-by-step solution

Here, for each x i = i , weight = w i = i 2 + i Hence, the required mean = Σ w i x i Σ w i = ∑ i = 1 9 i i 2 + i ∑ i = 1 9 i 2 + i = ∑ i = 1 9 i 3 + ∑ i = 1 9 i 2 ∑ i = 1 9 i 2 + ∑ i = 1 9 i = 9 2 n + 1 2 4 + 9 9 + 1 18 + 1 6 9 9 + 1 18 + 1 6 + 9 9 + 1 2 = 7

Practice Statistics on Quantrex Academy →

More from Statistics

Consider a data consisting of 10 observations x₁, x₂, , x₁₀ , whose mean is 5 and variance is 7 . If the mean and the variance of the first 8 observations x₁, x₂, , x₈ are 4 and 3. 2026 A set of four observations has mean 1 and variance 13 . Another set of six observations has mean 2 and variance 1 . Then, the variance of all these 10 observations is equal to: 2026 Let the mean and the variance of seven observations 2, 4, , 8, , 12, 14 , < , be 8 and 16 respectively. Then the quadratic equation whose roots are 3 + 2 and 2 + 1 is : 2026 A data consists of 20 observations x₁, x₂, , x₂₀ . If _ i=1 ²⁰(x_i + 5)^2 = 2500 and _ i=1 ²⁰(x_i - 5)^2 = 100 , then the ratio of mean to standard deviation of this data is: 2026 A variable X takes values 0, 0, 2, 6, 12, 20, , n(n-1) with frequencies ^nC₀, ^nC₁, ^nC₂, ^nC₃, ^nC₄, ^nC₅, , ^nC_n , respectively. If the mean of this data is 60 , then its median 2026 The mean deviation about the mean for the data x_i 5 7 9 10 12 15 f_i 8 6 2 2 2 6 is equal to: 2026 For 10 observations x₁, x₂, , x₁₀ , if _ i=1 ¹⁰(x_i+2)^2=180 and _ i=1 ¹⁰(x_i-1)^2=90 , then their standard deviation is: 2026 Suppose that the mean and median of the non-negative numbers 21, 8, 17, a, 51, 103, b, 13, 67, (a > b) , are 40 and 21 , respectively. If the mean deviation about the median is 26 2026 Full Statistics list All NTA Abhyas JEE Main PYQs