Question

The average weight of 64 students was calculatedas 46 kg. It was later found that the weight of one ofthe students was misread as 56 kg instead of 65 kg.What is the actual average weight of the group of64 students?​

Answer 1

$\huge \underline {\underline{ \mathfrak{ \green{Ans}wer \colon}}}$

As we know that :

$\large{\boxed{\sf{Avg \: = \: \dfrac{Total \: Weight}{No. \: of \: students}}}} \\ \\ \implies {\sf{46 \: = \: \dfrac{Total}{64}}} \\ \\ \implies {\sf{Total \: = \: 46 \: \times \: 64}} \\ \\ \implies {\sf{Total \: = \: 2944}} \\ \\ {\underline{\sf{\therefore \: Sum \: of \: Weight \: of \: students \: is \: 2944 \: kg}}}$

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Now, if weight of 1 student was missread as 56 instead of 65.

⇒Actual Weight = 2944 - 56 + 65

⇒Actual Weight = 2953 kg

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Now, use formula for average :

$\large{\boxed{\sf{Avg \: = \: \dfrac{Total \: weight}{No. \: of \: Students}}}} \\ \\ \implies {\sf{Avg \: = \: \dfrac{2953}{64}}} \\ \\ \implies {\sf{Avg \: = \: 46.14}} \\ \\ \underline{\sf{\therefore \: New \: Average \: Weight \: is \: 46.14 \: kg}}$

Answer 2

Answer:

Given:

Average Weight of 64 students is 46 kg. Later it was found that one of the students was misread as 56 kg instead of 65 kg.

To find:

Actual average of the 64 students.

Calculation:

Initial mistaken average is :

$\dfrac{x1 + x2 + ..... + 56}{64} = 46$

$= >( x1 + x2 + ..... + 56)= 46 \times 64$

$= >( x1 + x2 + ..... + 56)=2944$

But we know that actual weight of the student is 65 kg.

So new sum of weight will be :

$= ( x1 + x2 + ..... + 56) + 9$

$= 2944 + 9$

$= 2953 \: kg$

So new actual average will be :

$\therefore \: new \: average \: = \dfrac{2953}{64}$

$= > new \: average \: = 46.140625 \: kg$

So final answer :

$\boxed{ \large{ \sf{ \bold{ \green{new \: average \: = 46.140625 \: kg}}}}}$