Exams › JEE Advanced › Maths

An online examination is taken by 50 students, of whom 20 are boys. The boys have an average score of 12 with variance 2. The 30 girls also have a variance of 2. The overall average score of all 50 students is 15. If mu is the average score of the girls and sigma² is the variance of scores of all 50 students, find the value of mu + sigma².

1. 25
2. 20
3. 15
4. 30

Correct answer: 25

Solution

Overall mean: (20*12 + 30*mu)/50 = 15 => 240 + 30*mu = 750 => mu = 17. d1 = 12-15 = -3, d2 = 17-15 = 2. Combined variance = (20*(2+9) + 30*(2+4))/50 = (20*11 + 30*6)/50 = (220+180)/50 = 400/50 = 8. mu + sigma² = 17 + 8 = 25.

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 →