The mean and SD of the scoreing marks of 100 students are 40 and 5.1 resp. on checking it was discovered that the score which should correctly read as 40 had been wrongly taken as 50. Find the correct values of mean and SD

The mean and S.D. of 20 items were found to be 12 and 6 respectively. On checking it was discovered that items which should correctly read as 11 and 21 had been wrongly taken as 15 nd 27 respectively. Find the correct values of mean and S.D.

Mean of 100 items is 49. It was discovered that three items which should have been 60, 70,80 were wrongly read as 40, 20, 50 respectively. The correct mean is. (a) 48 (b) 82 1/2 (c) 50 (d) 80

For a group of 200 candidates the mean and S.D. were found to be 40 and 15 respectively. Later on it was found that the score 43 was misread as 34. Find the correct mean and correct S.D.

A student obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was later found that he had wrongly copied one observation as 50, the correct figure being 40. Calculate the correct mean and correct standard deviation.

For a group of 200 students the mean and S.D. of marks obtained by them were found to be 40 and 15 respectively. Later on, it was found that the score 23 was misread as 32. Find the correct mean and correct S.D.

For a group of 200 students the mean and S.D of marks obtained by them were found to be 40 and 15 respectively.Later on,it was found that the score 23 was misread as 32.Find the correct mean and correct S.D

The average mark obtained by 100 students is 40. But is was later found that the mark of one student was wrongly written as 83 instead of 53. Find the value of the actual mean.

The mean and SD of 20 items were found to be 12.5 and 2.1 resp.After that,one item of 23 being added: find the AM and SD of 21 items

The mean and standard deviation of 100 observations were calculated as 40 and 5.1, respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation?

The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation ?