Exams › JEE Main › Maths
If the mean and the standard deviation of 50 observations x1, x2,..., x50 are equal to 16, then the mean of (x1 - 4)², (x2 - 4)²,......, (x50 - 4)² is:
- 400
- 480
- 380
- 525
Correct answer: 400
Solution
The mean of the squared deviations from a constant can be calculated using the formula for variance. Since the mean of the original observations is 16, the mean of the transformed observations (x_i - 4)² results in a mean of 400, as it shifts the mean down by 4 and squares the result.
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 →