Exams › JEE Advanced › Maths
Given sum(i=1 to 10) of (x_i - 5) = 5 and sum(i=1 to 10) of (x_i - 5)² = 125, find the variance of x1, x2,..., x10.
- 9/4
- 25/4
- 49/4
- 81/4
Correct answer: 49/4
Solution
Since variance is shift-invariant, Var(x) = Var(x - 5). Mean of (x_i - 5) = 1/2, mean of (x_i - 5)² = 25/2. Variance = 25/2 - (1/2)² = 49/4.
Related JEE Advanced Maths questions
- The standard deviation of a set containing 25 values is 40. If 5 is added to each value in the set, what will the updated standard deviation be?
- Given Σ(xi − 5) = 9 and Σ(xi − 5)² = 45 for a dataset of 9 values x1, x2,..., x9, what is the standard deviation of the data?
- The variables X and U are connected by the equation X = 5 + 2U. If the mean of X is 10 and its coefficient of variation is 2.6, what is the coefficient of variation for U?
- What is the average absolute deviation from the mean for the arithmetic progression a, a + d, a + 2d,..., a + 2nd?
- What is the variance of the first n positive integers?
- Seven real numbers 9 = x1 < x2 <... < x7 form an arithmetic progression with common difference d. If their standard deviation is 4 and their mean is x_bar, find the value of x_bar + x6.
⚔️ Practice JEE Advanced Maths free + battle 1v1 →