The mean and S.D. of marks of 70 students were found to be 65 and 5.2 respectively. Laterit was detected that the marks of one student was wrongly recorded as 85 instead of 58.Obtain the correct s.d

Step 1: Calculate the total sum of all the marks. Since we know the mean (average) is 65 for 70 students, the total sum would be 65 * 70 = 4550.

Step 2: Correct the error in the data. The total sum of all the marks should have been 4550 - 85 (wrongly recorded mark) + 58 (correct mark) = 4523.

Step 3: Calculate the correct mean. The correct mean would be the total sum divided by the number of students, which is 4523 / 70 = 64.61 (rounded to two decimal places).

Step 4: Calculate the sum of the squares of the deviations from the mean. This is done by subtracting the mean from each individual mark, squaring the result, and then summing all these squares. However, since we only have the wrongly recorded mark and the standard deviation, we can use the formula for the standard deviation, which is sqrt[(sum of squares)/n - (mean)^2]. We can rearrange this to find the sum of squares = (standard deviation)^2 * n + n*(mean)^2. Substituting the given values, we get (5.2)^2 * 70 + 70*(65)^2 = 23684.

Step 5: Correct the sum of squares. The sum of squares should have been 23684 - 85^2 + 58^2 = 23425.

Step 6: Calculate the correct standard deviation. The correct standard deviation would be sqrt[(sum of squares)/n - (mean)^2]. Substituting the corrected values, we get sqrt[23425/70 - (64.61)^2] = 5.07 (rounded to two decimal places).

So, the correct standard deviation is 5.07.