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For the following data, find the weighted mean.​​Xi435721Wi102135\begin{array}{c|cccccc}X_i & 4 & 3 & 5 & 7 & 2 & 1 \\ W_i &am
Question

For the following data, find the weighted mean.

​​Xi435721Wi102135\begin{array}{c|cccccc}X_i & 4 & 3 & 5 & 7 & 2 & 1 \\ W_i & 1 & 0 & 2 & 1 & 3 & 5 \\ \end{array}​​

A.

2.66

B.

7.5

C.

5.6

D.

8

Correct option is A

Given:

Xi=[4,3,5,7,2,1], Wi=[1,0,2,1,3,5]X_i = [4, 3, 5, 7, 2, 1], \\ \ \\ W_i = [1, 0, 2, 1, 3, 5]​​

Formula Used:

WiWeighted Mean=(XiWi)Wi∑Wi\text{Weighted Mean} = \frac{\sum (X_i \cdot W_i)}{\sum W_i} ​​​

Solution:

Calculating XiWiX_i \cdot W_i​ ​ for each i:

[41, 30, 52, 71, 23, 15]=[4,0,10,7,6,5][4 \cdot 1, \, 3 \cdot 0, \, 5 \cdot 2, \, 7 \cdot 1, \, 2 \cdot 3, \, 1 \cdot 5] = [4, 0, 10, 7, 6, 5]​​

Sum of XiWi:X_i \cdot W_i​:​​

(XiWi)=4+0+10+7+6+5=32\sum (X_i \cdot W_i) = 4 + 0 + 10 + 7 + 6 + 5 = 32​​

Sum of Wi:W_i​:​​

Wi=1+0+2+1+3+5=12\sum W_i = 1 + 0 + 2 + 1 + 3 + 5 = 12​​

the weighted mean:

Weighted Mean=3212=2.67 (rounded to two decimal places)\text{Weighted Mean} = \frac{32}{12} = 2.67 \, (\text{rounded to two decimal places})​​

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