JEE Main202523 Jan 2025Evening ShiftMathematicsStatisticsActual

The variance of the numbers 8,21,34,47, , 320 is

Correct answer

8788

Step-by-step solution

aligned & 8+(n-1) 13=320 & 13 n=325 & n=25 & no. of terms =25 aligned aligned & mean = x _ i n = 8+21+ +320 25 = 25 2 (8+320) 25 & variance ^2= x _ i ^2 n -( mean )^2 & = 8^2+21^2+ .+320^2 13 -(164)^2 & =8788 aligned

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 JEE Main PYQs