Proof Prove that ∑i=1^n(xi-x̅)^2=∑i=1^n xi^2-(1)/(n)(∑i=1^n xi)^2

Name: Problem 103, Chapter 9: Precalculus
Uploaded: 2021-02-07T14:40:07-08:00
Duration: 5 min 11 s
Channel: Anurag Kumar
Description: Proof Prove that ∑i=1^n(xi-x̅)^2=∑i=1^n xi^2-(1)/(n)(∑i=1^n xi)^2

Proof Prove that ∑i=1^n(xi-x̅)^2=∑i=1^n xi^2-(1)/(n)(∑i=1^n...

Key Concepts

Algebraic Expansion

This concept involves expanding the square of a difference, as in (x - ?)^2, by applying the distributive property (FOIL method). Expanding the squared term allows us to break it into individual components that can be manipulated and combined with similar terms, which is essential in deriving various formulas in algebra and statistics.

Properties of Summations

This concept pertains to the rules that govern the operation of summing sequences, including linearity, which allows the summation of a sum to be split into the sum of individual summations, as well as handling constant multipliers. These properties enable the separation and recombination of terms to simplify complex expressions involving series.

Relationship Between Total Sum of Squares and Variance

This concept connects the idea of measuring dispersion by comparing the sum of squared deviations from the mean with the sum of the squares of the data points. It forms the basis for defining variance, showing that the total variation can be expressed as the sum of squares of individual observations reduced by a correction factor related to the mean, highlighting the error adjustment needed when computing variability.

Transcript

00:01 In this problem of here series we have to prove that this is sigma i equals to this is 1 and here 2n and this is x i minus x bar this is the arithmetic mean whole square is equal to this is sigma i equals to 1 to n and this is x i equal to n and this is x i whole square minus 1 divided with n and this is sigma again here sigma i equals to 1 to this is n and this is x i and whole square and now we have to prove that so we know that here x bar is equal to 1 divide with n from here i equals to 1 to n x i so this is the value of x bar so simply do the squaring from the the left hand side.

01:03 So we are starting from, starting from left hand side.

01:11 So this would be here, sigma i equals to 1 n x i minus x bar whole square.

01:21 So simply do the whole squaring i equals to 1 to n so this would be here x i whole square plus x bar whole square minus 2 x i x bar.

01:33 So here break into parts.

01:37 So this would be here from i equals to 1 to n and x i this would be whole square and here x bar is taken common so this would be here summation of this is here simply this is summation of this is here summation of i is equal to 1 to n and this would be here x bar whole square minus 2 x i x i x bar so now take here x bar common so here so here this would be like same i equals to 1 to infinity and x i whole square and here take x bar as common so we left with here x bar minus 2x i so here this is the 2x i and now we have here so simply multiply n and divide n so multiply and then divide n.

02:44 So this would be here i is equal to 1 to n and x i whole square.

02:50 And here it would be summation i equals to 1 to n and send it to the inside and then here it would be 1 divide with an outside...

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