Step 1: Recall definition of variance for Gaussian random variables.
For a Gaussian random variable with mean zero and standard deviation $\sigma_{0}$,
\[
E[x_{n}^{2}] = \sigma_{0}^{2}.
\]
Step 2: Expectation of $\hat{\sigma_{1}^{2}$.}
\[
\hat{\sigma}_{1}^{2} = \frac{1}{10000}\sum_{n=1}^{10000} x_{n}^{2}.
\]
Taking expectation:
\[
E[\hat{\sigma}_{1}^{2}] = \frac{1}{10000}\sum_{n=1}^{10000} E[x_{n}^{2}]
= \frac{1}{10000} \times 10000 \times \sigma_{0}^{2}
= \sigma_{0}^{2}.
\]
Thus, $\hat{\sigma}_{1}^{2}$ is an unbiased estimator of variance.
Step 3: Expectation of $\hat{\sigma_{2}^{2}$.}
\[
\hat{\sigma}_{2}^{2} = \frac{1}{9999}\sum_{n=1}^{10000} x_{n}^{2}.
\]
Taking expectation:
\[
E[\hat{\sigma}_{2}^{2}] = \frac{1}{9999} \times 10000 \times \sigma_{0}^{2}
= \frac{10000}{9999}\sigma_{0}^{2}.
\]
This is slightly larger than $\sigma_{0}^{2}$, so $\hat{\sigma}_{2}^{2}$ is a biased estimator.
Step 4: Check the given options.
- (A) False, since $E[\hat{\sigma}_{2}^{2}] \neq \sigma_{0}^{2}$.
- (B) False, since expectation of $\hat{\sigma}_{2}$ (square root form) is not equal to $\sigma_{0}$.
- (C) True, because $E[\hat{\sigma}_{1}^{2}] = \sigma_{0}^{2}$.
- (D) False, because $\hat{\sigma}_{1}$ and $\hat{\sigma}_{2}$ are computed differently.
% Final Answer
\[
\boxed{\text{Option (C): } E(\hat{\sigma}_{1}^{2}) = \sigma_{0}^{2}}
\]
Let the mean and variance of 7 observations 2, 4, 10, x, 12, 14, y, where x>y, be 8 and 16 respectively. Two numbers are chosen from \(\{1, 2, 3, x-4, y, 5\}\) one after another without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4, is:
If the mean and the variance of the data 
are $\mu$ and 19 respectively, then the value of $\lambda + \mu$ is
A continuous time periodic signal \( x(t) \) is given by: \[ x(t) = 1 + 2\cos(2\pi t) + 2\cos(4\pi t) + 2\cos(6\pi t) \] If \( T \) is the period of \( x(t) \), then evaluate: \[ \frac{1}{T} \int_0^T |x(t)|^2 \, dt \quad {(round off to the nearest integer).} \]
The maximum percentage error in the equivalent resistance of two parallel connected resistors of 100 \( \Omega \) and 900 \( \Omega \), with each having a maximum 5% error, is: \[ {(round off to nearest integer value).} \]
Consider a distribution feeder, with \( R/X \) ratio of 5. At the receiving end, a 350 kVA load is connected. The maximum voltage drop will occur from the sending end to the receiving end, when the power factor of the load is: \[ {(round off to three decimal places).} \]
In the circuit with ideal devices, the power MOSFET is operated with a duty cycle of 0.4 in a switching cycle with \( I = 10 \, {A} \) and \( V = 15 \, {V} \). The power delivered by the current source, in W, is: \[ {(round off to the nearest integer).} \] 
The induced emf in a 3.3 kV, 4-pole, 3-phase star-connected synchronous motor is considered to be equal and in phase with the terminal voltage under no-load condition. On application of a mechanical load, the induced emf phasor is deflected by an angle of \( 2^\circ \) mechanical with respect to the terminal voltage phasor. If the synchronous reactance is \( 2 \, \Omega \), and stator resistance is negligible, then the motor armature current magnitude, in amperes, during loaded condition is closest to: \[ {(round off to two decimal places).} \]