Exams › JEE Advanced › Maths

Let alpha and beta be real numbers. Six observations are -3, 4, 7, -6, alpha, beta with mean 2 and variance 23. Find the mean deviation about the mean of these six observations.

1. 13/3
2. 16/3
3. 11/3
4. 14/3

Correct answer: 13/3

Solution

Mean = 2: (-3+4+7-6+alpha+beta)/6 = 2 => 2+alpha+beta = 12 => alpha+beta = 10. Variance = 23: sum of (xi-2)²/6 = 23. (xi-2)² for known values: 25+4+25+64 = 118. So (alpha-2)²+(beta-2)² = 138-118 = 20. With alpha+beta=10: (alpha²+beta²) - 4*10 + 8 = 20 => alpha²+beta² = 52. Then 2*alpha*beta = (alpha+beta)² - (alpha²+beta²) = 100-52 = 48 => alpha*beta=24. So alpha=6, beta=4. Observations: -3,4,7,-6,6,4. Deviations from mean 2: 5,2,5,8,4,2. Mean deviation = (5+2+5+8+4+2)/6 = 26/6 = 13/3.

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 →