Exams › JEE Advanced › Maths

For five observations x1, x2, x3, x4, x5, it is given that the sum of (xi - 10) for i = 1 to 5 equals 5, and the sum of (xi - 10)² for i = 1 to 5 equals 25. Find the standard deviation of the transformed observations 2*x1 + 7, 2*x2 + 7, 2*x3 + 7, 2*x4 + 7, 2*x5 + 7.

1. 8
2. 16
3. 4
4. 2

Correct answer: 4

Solution

Let ai = xi - 10. We have sum(ai) = 5, sum(ai²) = 25. Mean of ai = 5/5 = 1. Variance of ai = (25/5) - 1² = 5 - 1 = 4, so SD(xi) = 2. The new observations yi = 2*xi + 7 are a linear transformation. SD(yi) = |2| * SD(xi) = 2 * 2 = 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 →