Question

The average marks of class of 45 students is 36 . Later it was realized that two students whose marks were 30 and 54 were wrongly recorded 13 and 45 . Find the correct mean

Answer 1

$\small{ \sf{Average \: marks \: of \: class \: of \: 45 \: students = 36}} \\ \small{ \sf{Sum \: of \: marks \: of \: these \: students = 45 \times 36}} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1620$

$\green{ \tiny \bf{Correct \: Sum = [Incorrect \: sum - (Sum \: of \: wrong \: records)] + (Sum \: of \: right \: records)</p><p>} }\\ \bf{Correct \: Sum = </p><p>[1620 - (13 + 45)</p><p>] + (30 + 54)} \\ \\ \: \: \: \: \: \implies[1620 - 58] + 84 \\ \: \: \: \: \: \implies1620 - 58 + 84 \\ \: \: \: \: \: \implies1620 + 26 = 1646$

Hence,

$\tt{Correct \: \: Mean = \frac{1646}{45} = \red{36.578} }$