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The mean thickness and variance of silicon steel laminations are 0.2 mm and 0.02 respectively. The varnish insulation is applied on both the sides of the laminations. The mean thickness of one side insulation and its variance are 0.1 mm and 0.01 respectively. If the transformer core is made using 100 such varnish coated laminations, the mean thickness and variance of the core respectively are

R S Selvaraj
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  • Welcome to mathSE! What have you tried – rlartiga May 15 '14 at 20:30
  • I have found the mean thickness to be 100*(0.1+0.2+0.1) which is 40mm. – R S Selvaraj May 16 '14 at 06:27
  • I have found the mean thickness to be 100(0.1+0.2+0.1) which is 40mm.But while trying to get variance, for one varnish coated lamination i got the variance to be 2^2(0.01) + 0.02 = 0.06 but i am not sure whether it is correct. Moreover, for the transformer core, i cannot multiply this 0.06 by 100 as the random variables involved need not be independent. Thus, i am stuck-up with calculating variance. – R S Selvaraj May 16 '14 at 06:39

2 Answers2

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If $X_i$ is the thickness of the lamination $i, \ i=1 \cdots 100$ each one with a normal distribution with mean $\mu_x=0.2$ and variance $\sigma^2_x=0.02$.

For each lamination is two varnish insulations therefore there is $200$ insulations, denote by $Y_j$ the thickness of the insulation $j, \ j=1 \cdots 200$ each one with a normal distribution with mean $\mu_y=0.1$ and variance $\sigma^2_y=0.01$.

Hence the thickness of the transformer is:

$$T=\sum_{i=1}^{100} X_i + \sum_{j=1}^{200} Y_j $$

Assuming the variables $X_i,Y_j$ are independent each other:

$$E(T)=\sum_{i=1}^{100} E(X_i) + \sum_{j=1}^{200} E(Y_j)=100\mu_x + 200 \mu_y $$ $$Var(T)=Var\left(\sum_{i=1}^{100} X_i + \sum_{j=1}^{200} Y_j \right)=\sum_{i=1}^{100} Var(X_i) + \sum_{j=1}^{200} Var(Y_j)=100\sigma^2_x+200 \sigma_y^2$$

rlartiga
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  • Regarding mean, i agree with your answer 100(0.2) + 200(0.1) = 40mm. But regarding variance, i cannot agree with you - for, X_i are equal and cannot be considered to be independent and moreover, Var(X + X) is not equal to Var(X) + Var(X); it is 2^2Var(X). – R S Selvaraj May 17 '14 at 16:07
  • Perhaps you may want to look this: http://en.wikipedia.org/wiki/Variance#Sum_of_uncorrelated_variables_.28Bienaym.C3.A9_formula.29 and second one thing they are equally distributed and another thing is there are not independent. – rlartiga May 18 '14 at 01:56
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this my solution (and i am not 100% sure) there is a mistake in the question . It should be standard deviation in the place of variance.This problem is similar to the question in the book "Advance engg mathematics" by erwin krezig in the sec 24.3 ,problem 10 (or 9 depending on the edition). according to a theorem in two dimensional random variable "the variance of random variable Z=X+Y is equal to sum of variance of random variables X and Y if X and Y are independant"

going by this theorem and considering the metnioned values are standard deviation instead of variance.the variance of transformer = 100*(.02^2)+200*(.01^2)= 0.06 and standard deviation is sqrt of variance = 0.24 mm. This corresponds to the option D in question ,which is also given in the kEY to this question paper.

avk
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