2. Let f = xz - yz, V⃗ = 4zi + 2yj + (x

2. Let $f = xz - yz$, $\vec{V} = 4zi + 2yj + (x - z)k$, $\vec{W} = y^2i + (y^2 - x^2)j + 2z^2k$. Find: i) $\vec{\nabla} \cdot (\vec{\nabla}f)$ ii) $\nabla^2(xzf)$ iii) $\vec{V} \cdot (\vec{\nabla} \times \vec{W})$

Transcript

00:01 Hello students in the first question we have given that the random variable x1 x2 xn follows the exponential distribution with parameter theta.

00:10 So that means we have xi follows exponential distribution with parameter theta and i is from 1 to n then by using that we have to derive 1 minus alpha into 100 % approximate confidence interval for the parameter theta.

00:28 Now the pdf of x is given by f of x that is equal to theta into e raised to minus theta x and x is greater than 0.

00:39 Now by using this we can find the likelihood function that is l which is the product of the pdf.

00:48 So product i from 1 to n f of xi and that is equal to theta raised to n into e raised to minus theta into summation of xi.

01:01 Now here by taking the partial differentiation with respect to theta that is daba of daba theta of log of l is equal to daba of daba theta of n of log of theta minus theta of summation of xi.

01:21 Now this is equal to n into daba of daba theta of log of theta is 1 by theta.

01:27 So this is n by theta minus daba of daba theta of theta is 1.

01:33 So this is only summation of xi and that is equal to n divided by theta minus n into x bar which is equal to n into bracket 1 divided by theta minus x bar.

01:48 Now again differentiating this with respect to theta.

01:52 So that means we are taking the double differentiation.

01:56 So this is minus daba square divided by daba theta square log of l is equal to n divided by theta square and which implies variance of daba of daba theta of log of l and that is equal to expected value of minus daba square divided by daba theta square of log of l and that is equal to n divided by theta square.

02:30 Now, hence for the large samples, we will get hence for large samples z that is the standard normal variable is equal to n into 1 divided by theta minus x bar divided by square root of n divided by theta square and it follows the standard normal distribution.

03:00 That means normal with parameter 0 comma 1 and this implies square root of n into 1 minus theta x bar.

03:09 It follows normal 0, 1.

03:13 Hence 95 % confidence limits or confidence interval for theta is given by so this is probability of minus 1 .96 is less than or equal to square root of n into 1 minus theta x bar and that is less than or equal to 1 .96 which is equal to 0 .95 then square root of n into 1 minus theta x bar.

03:44 It is less than or equal to 1 .96 and that implies 1 minus 1 .96 divided by square root of n into 1 divided by x bar and that is less than or equal to theta...

Question

2. Let $f = xz - yz$, $\vec{V} = 4zi + 2yj + (x - z)k$, $\vec{W} = y^2i + (y^2 - x^2)j + 2z^2k$. Find: i) $\vec{\nabla} \cdot (\vec{\nabla}f)$ ii) $\nabla^2(xzf)$ iii) $\vec{V} \cdot (\vec{\nabla} \times \vec{W})$

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