Exams › JEE Advanced › Maths

A frequency distribution has values x = c, 2c, 3c, 4c, 5c, 6c with corresponding frequencies f = 2, 1, 1, 1, 1, 1 respectively. If the variance of this distribution is 160, find the natural number c.

1. 5
2. 8
3. 7
4. 6

Correct answer: 7

Solution

With total frequency n = 7, the mean works out to 22c/7. Computing the second moment and subtracting the mean squared gives variance = 160c²/49. Setting this equal to 160 yields c = 7.

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 →