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  • The mean and standard deviation of the marks of 200 candidates were found to be 40 and 15, respectively. Later, it was discovered that a score of 40 was wrongly read as 50. What are the correct mea

The mean and standard deviation of the marks of 200 candidates were found to be 40 and 15, respectively. Later, it was discovered that a score of 40 was wrongly read as 50. What are the correct mean and standard deviation, respectively?

 

Option: 1

40 and 15


Option: 2

40 and 14


Option: 3

48 and 14


Option: 4

39 and 14


Answers (1)

To find the correct mean and standard deviation, we need to adjust the wrongly read score and recalculate these statistics.

First, let's adjust the score of 50 to the correct score of 40.

The sum of the original 200 scores is 40 \times 200 = 8000.

We need to subtract the wrongly read score (50) and add the correct score (40) to the sum.

Corrected sum = (8000 - 50) + 40 = 8000 - 50 + 40 = 8010.

The correct mean is the corrected sum divided by the number of candidates (200).

Corrected mean = 8010 / 200 = 40.05.

Next, let's adjust the standard deviation. Since we only changed one score, the impact on the standard deviation is minimal.

The corrected standard deviation remains the same as the original standard deviation, which is 15.

Therefore, the correct mean is approximately 40.05, and the correct standard deviation is 15.

Among the given options, 40 and 14 is the closest to the correct mean and standard deviation.

 

Posted by

Sumit Saini

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