Average scores of fifty students in a class is $44$ . Later on it was found that a score $23$ was incorrectly recorded as $73$ . The correct average score is

42

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43

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45

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44

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Solution

The correct option is B $43$
Average of $50$ students $= 44$ .
Sum of marks of $50$ students = $44 \times 50$ $= 2200$ .
After correcting the score of $73$ to $23$ , new sum $= 2200 - 73 + 23 = 2150$
Mean = $\frac{2150}{50}$
$= 43$

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Similar questions

Q. The average marks of 100 students were found to be 40. Later on it was discovered that a score of 53 was misread as 83. Find the correct average corresponding to the correct score.

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The mean of the marks scored by 50 students was found to be 39. Later on it was discovered that a score of 43 was misread as 23. The correct mean is
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In a Mock CAT, 123 students appeared and the average score obtained was 120. But later it was found that the top three students were repeaters, so their score has been eliminated and then the new average score was found to be decreased by 1.5. Also, it is known that all the students obtained the marks in integers and the scores of the toppers were distinct. If the second highest topper has scored more than 185 marks, then the highest possible score of the third highest topper is :

Q. The mean mark of

100

students was found to be

40

. Later on, it was found that a score of

53

was misread as

83

. Find the correct mean corresponding to the correct score.

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Average scores of fifty students in a class is 44. Later on it was found that a score 23 was incorrectly recorded as 73. The correct average score is

The correct option is B 43Average of 50 students =44.Sum of marks of 50 students = 44×50 =2200.After correcting the score of 73 to 23, new sum =2200−73+23=2150Mean = 215050=43

Average scores of fifty students in a class is $44$ . Later on it was found that a score $23$ was incorrectly recorded as $73$ . The correct average score is

The correct option is B $43$
Average of $50$ students $= 44$ .
Sum of marks of $50$ students = $44 \times 50$ $= 2200$ .
After correcting the score of $73$ to $23$ , new sum $= 2200 - 73 + 23 = 2150$
Mean = $\frac{2150}{50}$
$= 43$