Question (7.10): (7.12): 7.8. ** The weighted average (7.10) of N separate measurements is a simple function of x1, x2, ..., xy. Therefore, the uncertainty in Xwav can be found by error propagation. Prove in this way that the uncertainty in xway is as claimed in (7.12). Ew;x; Ew;? (7.10) where the sums are over all N measurements, i = 1, ...,N, and the weight w; of each measurement is the reciprocal square of the corresponding uncertainty, (7.11) for i = 1, 2, ..., N. This docu Because the weighted average Xway, is a function of the original measured values X1, X2, ..., Xy, the uncertainty in Xway can be calculated using error propagation. As you can easily check (Problem 7.8), the uncertainty in Xway is (7.12) Oway =

Answer: The weighted average of N separate measurements is a simple function of x_{1}, x_{2}, \ldots x_{N}. Therefore, the uncertainty in x wow can be found by error propagation.To prove: The uncertainty in x waw is clcimed.Now,consider the weighted average of N separate measurements x, x_{2}, \ldots x, x_{N}, with uncer tainties \sigma_{1}, \sigma_{2}, \cdots \sigma_{N}.\begin{array}{l}\operatorname{Prob} x\left(x_{A}\right) 2 \frac{1}{\sigma_{B}} e^{-\left(x_{A}-x\right)^{2} / 2 \sigma_{A}^{2}}(1) \\\operatorname{Prob} x\left(x_{B}\right) 2 \frac{1}{\sigma_{B}} e^{\frac{\left(x_{B}-x\right)^{2}}{2 \sigma_{B}}}-(2) \\\operatorname{Prob}\left(x_{A}, x_{B}\right)=\operatorname{Probx}\left(x_{A}\right) \cdot \operatorname{Prob}\left(x_{B}\right) \\2 \frac{1}{\sigma_{A} \sigma_{B}} e^{\frac{-x^{2}}{2}}\end{array}Here,I is the convenient shorthand.Nowx^{2} (chisquared) for the exponent.\begin{array}{l}x^{2}=\left(\frac{x_{A}-x}{\sigma_{A}}\right)^{2}+\left(\frac{x_{B}-x}{\sigma_{B}}\right)^{2} \cdot \frac{2 x_{A}-x}{\sigma_{A}^{2}}+\frac{2 x_{B}-x}{\sigma_{B}^{2}}=0 \\x=\frac{\left(\frac{x_{A}}{\sigma^{2} A}+\frac{x_{B}}{\sigma^{2} B}\right)}{\left(\frac{1}{\sigma^{2} A}+\frac{1}{\sigma_{B}}\right)}\end{array} \therefore W_{A}=\frac{1}{\sigma_{A}^{2}}andW_{B}=\frac{1}{\sigma^{2} B}\therefore x_{\text {wav }}=\frac{W_{A} X_{A}+W_{B} X_{B}}{W_{A}+W_{B}}Weighted awerage isx_{\text {aur }}=\sum_{i=1}^{N} \frac{w_{i} x_{i}}{\sum_{i=1}^{N} w_{i}}Now the weights arew i=\frac{1}{\sigma_{i}^{2}}The uncertainty on xavg\sigma_{\operatorname{ang}}=\frac{1}{\sqrt{\sum_{i=1}^{N} w_{i}}}If \sigma_{1}=\sigma_{2}=\sigma_{3}=\cdots \cdots \sigma_{N}=\sigma_{\text {, }}then the weighted werage is the simple average.The uncertainty is \frac{\sigma}{\sqrt{n}} and theweighted average have small uncentainty. which goes into average.Hence proved. NOTE::As per the rules i can answered ONLY ONE QUESTION. I HOPE YOUR HAPPY WITH MY ANSWER....***PLEASE SUPPORT ME WITH YOUR RATING... ***PLEASE GIVE ME "LIKE"...ITS VERY IMPORTANT FOR ME NOW....PLEASE SUPPORT ME ....THANK YOU ...