Exams › JEE Main › Maths
The mean square deviation of n observations about a point c is (1/n) * sum (xi - c)². If the mean square deviations about -2 and about 2 are 18 and 10 respectively, find the standard deviation of the data.
- 3
- 2
- 1
- None of these
Correct answer: 3
Solution
Using MSD(c) = sigma² + (xbar - c)²: subtracting gives xbar = 1, then sigma² = 9, so sigma = 3.
Related JEE Main Maths questions
- A student took an exam consisting of 5 subjects. In four of the subjects, he scored 90, 70, 75, and 65 marks. What should be the minimum and maximum marks in the fifth subject so that his overall average is at least 70 and at most 75?
- For a data set of 15 values of X, the given totals are Σx² = 2830 and Σx = 170. Later, one entry recorded as 20 was discovered to be incorrect and was changed to the correct value 30. The variance after this correction is
- What is the quartile deviation for the data set 12, 7, 15, 10, 16, 17, 25?
- Take the first ten positive integers. If each number is first multiplied by -1 and then increased by 1, what is the variance of the resulting set of numbers?
- For the grouped data below, what is the standard deviation of the distribution?
Class intervals: 0–10, 10–20, 20–30, 30–40
Frequencies: 1, 3, 4, 2
- Two distributions have coefficients of variation 50 and 60, and their arithmetic means are 30 and 25, respectively. What is the difference between their standard deviations?
⚔️ Practice JEE Main Maths free + battle 1v1 →