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The selling amount of money of a shopkeeper in 7 days of any week are given below (in rupees). 115,98, 102,126,85,91,107 Find the average of the amount of rupees sold per day. |
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Answer» Here, the values of the variable x , is near the values 100 of x If we transfer the origin A to 100 , then the values of u are 15-2,2,26,-15,-8,7. `therefore baru=1/7(15-2+2+26-15-9+7)=(50-26)/(7)=(24)/(7)=3.43` (approx.) `therefore baru=100+3.43 =103.43` (approx) `therefore ` the required daily seling amount =Rs. 103.43 Now , let x is a descrete variable. The possible values of `x_(1), x_(2),.............., x_(n)` . Frequency distribution of x are made from n values of x. Let the frequencies of `x_(1), x_(2), .........x_(n)` be `f_(1),f_(2), ..............f_(n)` i.e. `x_(1)` occurs `f_(1)` times , `x_(2)` occurs `f_(2)` times, .......`f_(n)` occurs n-times and `f_(1)+f_(2) +........+f_(n)=n` Now , `underset(i=1)overset(n)sum f_(i)x_(i)=x_(i)xxf_(i)+x_(2)xxf_(2) + ........+x_(n)xxf_(n)` and `=f_(1)+f_(2)+............+f_(n)=n` Arithmetic mean of `x=barx=(x_(1)f_(1)+x_(2)f_(2)+.....+x_(n)f_(n))/(f_(1)+f_(2)+.......+f_(n))` `=(underset(i=1)overset(n)sumx_(i)f_(i))/(underset(i=1)overset(n)sum f_(i)=(1)/(n)underset(i=1)overset(n)sum x_(i)f_(i)................(""^(**)4)` This `barx` is the weighted mean of the variable x. |
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