Statistical Averages MCQ Quiz in తెలుగు - Objective Question with Answer for Statistical Averages - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

Total power = E[X²] = R_X (0)

D.C. Power = [E[X]]² = R_X(∞)

Calculation:

We know, Variance = E[X²] – [E[X]]²

∵ E[X²] = R_X (0) = 30

[E[X]]² = R_X(∞) = 5

∴ Variance = 30 – 5

 = 25

Standard Deviation,

 ${\sigma }_X=\sqrt{Variance}=\sqrt{25}=5$

Concept:

For the discrete system 

The mean of output is given by

m_y = σ_x ∑h_i(n)

And the variance of output is given by

σ²y = σ²x ∑ h_i²(n)

Analysis:

Given that

mx = 2 and σ2x = 3

mean of output is given by

$m_y=2\times (\frac{1}{3}+\frac{1}{3}+\frac{1}{3})$

m_y = 2

The variance of output is given by

${\sigma }_y=3\times (\frac{1}{9}+\frac{1}{9}+\frac{1}{9})$

σ2y = 1

Hence option (1) is correct

Concept:

The mean of a random variable also called the expected value is defined as:

$E\left[X\right]=\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}x.f_x\left(x\right)dx$

$E\left[X^2\right]=\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}{x}^2{f}_x\left(x\right)dx$

Properties of mean:

1) E[K] = K, Where K is some constant

2) E[c X] = c. E[X], Where c is some constant

3) E[a X + b] = a E[X] + b, Where a and b are constants

4) E[X + Y] = E[X] + E[Y]

Application:

E[(2x + 1)²] = E(4x² + 4x + 1)

= 4E(x²) + 4E(x) + 1

$E\left(x\right)={\sum }_{i=1}^3{x}_{i}P\left(x_i\right)$

$=3\times \frac{1}{6}+6\times \frac{1}{2}+9\times \frac{1}{3}$

$=\frac{13}{2}$

$E\left(x^2\right)={\sum }_{i=1}^3{x}_i^{2}P\left(x_i\right)$

$=9\times \frac{1}{6}+36\times \frac{1}{2}+81\times \frac{1}{3}$

$=\frac{3}{2}+18+27$

$=\frac{93}{2}$

Now:

$E\left[{\left(2x+1\right)}^2\right]=4\times \frac{93}{2}+4\times \frac{13}{2}+1$

= 186 + 26 + 1

= 213

The probability density function f_X (x) is given below:

Since the magnitude of f_X(x) at x = 4 is not shown in figure, we assume it to be k.

i.e. f_X(x = 4) = k

Now, from the property of probability density function, we have

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{f}_X\left(x\right)dx=1$ 

$\underset{0}{\overset{4}{\int }}{f}_X\left(x\right)dx=1$       ---(1)

From the figure, we define the probability density function as

$\frac{f_X\left(x\right)-0}{x-0}=\frac{k-0}{4-0}$ 

$f_X\left(x\right)=\frac{k}{4}x,0<x\le 4$       ---(2)

Substituting it in equation (1), we obtain

$\underset{0}{\overset{4}{\int }}\frac{k}{4}xdx=1$ 

$\frac{k}{4}{\left[\frac{x^2}{2}\right]}_0^4=1$ 

$\frac{k}{4}\left[\frac{4^2}{2}-0\right]=1$ 

$k=\frac{1}{2}$ 

So, from equation (2) we get

$f_X\left(x\right)=\{\begin{array}{c}\frac{x}{x},0<x\le 4\\ 0,otherwise\end{array}$ 

Thus, the mean square value of a random variable is given by

${\overline{X}}^2=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2{f}_X\left(x\right)dx$ 

$\underset{0}{\overset{4}{\int }}{x}^2\left(\frac{x}{8}\right)dx={\left[\frac{x^4}{32}\right]}_0^4=8$ 

Concept:

Mean = Expected value

$Correctioncoefficient=\frac{Covariance}{\sqrt{{\sigma }_x^2{\sigma }_y^2}}$

Where, covariance of random variable X and y is defined as:

$COV\left(x,y\right)=E\left[\left(X-E\left[X\right]\right)\left(Y-E\left[Y\right]\right)\right]$

For given R.V

Let, X = Q

Y = R

E[Q] = E[R] = 0

$COV\left(Q,R\right)=E\left[QR\right]$

$Correlation=\frac{E\left[QR\right]}{\sqrt{4\times 16}}=-0.5$

Calculation:

E [Q, R] = -4

Mean, 

= E [a²Q² + 9R² + 6a QR]

⇒ a² E[Q²] + 9E[R²] + 6a E[QR]       ----(1)

We know,

$\left[\begin{array}{c}{\sigma }_X^2=E\left[X^2\right]-{\left(E\left[X\right]\right)}^2\\ E\left[X^2\right]={\sigma }_X^2+{\left(E\left[X\right]\right)}^2\end{array}\right]$

Since mean of both random variable Q and R are 0

E[Q] = E[R] = 0

$E\left[Q^2\right]={\sigma }_Q^2=4$

$E\left[R^2\right]={\sigma }_R^2=16$

Substitute in (1)

Mean [P] = a²(4) + 9(16) + 6a(-4)

⇒ 4a² + 144 - 24 a

To find minimum value differentiate

8a - 24 = 0

a = 3

Concept:

If ‘X’ is said to be normal Random Variable defined in - ∞ to ∞ with mean ‘μ’ and variance ‘ σ² ‘ then the Random variable is known as “Normal Random Variable”.

Its probability density function is defined as:

$N\left(X;\mu ,{\sigma }^2\right)=f\left(x\right)=\left\{\begin{array}{c}\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}(\frac{{\left(x-\mu \right)}^2}{{\sigma }^2}}\\ -\mathrm{\infty }<x<\mathrm{\infty }\\ -\mathrm{\infty }<\mu <\mathrm{\infty }\\ 0<\sigma <\mathrm{\infty }\\ 0;else\end{array}\right\}$

Variance = σ2

Calculation:

From the above equation comparing with the question we get

2σ² = 8

σ² = 4

The expected cost of machine A is

E (C_A) = E (160 + 40.X²)

= 160 + 40 E (X²)

$\begin{array}{l}=16040\left({\sigma }_x^2+{\left[E\left(X\right)\right]}^2\right)\\ \left(\therefore {\sigma }_x^2=E{\left(X\right)}^2-{\left[E\left(X\right)\right]}^2\right)\end{array}$

= 160 + 40 (0.96 + (0.96)²)

Since for a possion random variable mean is equal to variance = 235.26

The expected cost of machine B is

E (C_B) = E (128 + 40.Y²)

= 128 + 40 E (Y²)

$=128+40\left({\sigma }_y^2+{\left[E\left(Y\right)\right]}^2\right)$

= 128 + 40 (1.12 + (1.12)²)

= 223

The expected cost of machine A is 12.26 greater than machine B.

Concept:

Linearity: It is combination of superposition (additivity) and scaling (Homogeneity)

Additivity:

x₁(t) → y₁(t)

x₂(t) → y₂(t)

x₁(t) + x₂(t) → y₁(t) + y₂(t)

Scaling:

α x(f) → α y(t)

x(t) is input signal whereas y(t) is output signal.

Time invariant:

The system is time-invariant (T.I) if the input and output characteristics don’t change with time.

The time-varying nature is caused due to internal components.

T.I (Shift invariant) & T.V (shift dependent system)

If x(t) → y(t) then

X(t – t₀) → y(t – t₀)

$T.Iy\left(t\right){|}_{x\left(t-t_0\right)}=y\left(t\right){|}_{t=\left(t-t_0\right)}$

Calculation:

Given system is y[n] = x[-n]

Checking for time invariance:

${y\left[n\right]|}_{x\left[n-n_0\right]}$, i.e. shifting the signal by n_o, the result will be:

$y\left[n\right]{|}_{x\left[x-n_0\right]}=x\left[-n-n_0\right]$      ---(1)

Now substitute n – n₀ in place of n.

$y\left[n\right]{|}_{n=n-n_0}=x\left[-\left(n-n_0\right)\right]$

= x[-n + n₀]        ---(2)

Equation (1) and (2) are not the same, ∴  it is a time variant system.

Checking for linearity:

Additivity:

x₁[n] has output response as y₁[n]

x₂[n] has output response as y₂[n]

x₁[-n] → y₁[n]

x₂[-n] → y₂[n]

x₁[-n] + x₂[-n] → y₁[n] + y₂[n]

Scaling also satisfies.

So the given system is linear and Time variance system.

Important Conclusions:

1) Whenever checking for linearity, see only the time dependency.

y[n] = x[-n]

both are the same factor, so linear.

2) If the output is a trigonometric function of input, then the system will be nonlinear.

Example: y(t) = cos (x(t)) and y(t) = sample {x(t)}

3) Addition of constant makes system non-linear, i.e.

y(t) = a x(t) + b is non-linear

4) Division and multiplication of signal with an unknown signal also makes the system non linear.

5) Truncation/Rounding is “Quantization” which is non-linear. 

Concept:

1) (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac

2) E[a + b] = E[a] + E[b]

3) The autocorrelation function depends only on the time difference

Calculations:

$E{\left[\left(x\left(1\right)+x\left(2\right)+x\left(3\right)\right)\right]}^2$

$=E\left[x{\left(1\right)}^2+x{\left(2\right)}^2+x{\left(3\right)}^2+2x\left(1\right)x\left(2\right)+2x\left(1\right)x\left(3\right)+2x\left(2\right)x\left(3\right)\right]$       ----(1)

$=E\left[x{\left(1\right)}^2\right]=E\left[x{\left(2\right)}^2\right]=E\left[x{\left(3\right)}^2\right]=Rx\left(0\right)$

$E\left[x\left(1\right)x\left(2\right)\right]=R_x\left(-1\right)$

$E\left[x\left(1\right)x\left(3\right)\right]=R_x\left(-2\right)$ (Autocorrelation depends only on the time difference)

$E\left[x\left(2\right)x\left(3\right)\right]=R_x\left(-1\right)$

Equation (1) can be written as

$E\left[{\left(x\left(1\right)+x\left(2\right)+x\left(3\right)\right)}^2\right]$

3R_x (0) + 2 R_x(-1) + 2R_x(-2) + 2R_x(-1)

= 3×3 + 2×2 + 2×1 + 2×2

Concept:

Variance of random variable X is given by

${\sigma }_x^2=E\left[X^2\right]-{\left[E\left[X\right]\right]}^2$

here, X = cosθ

where, -π < θ < π

Calculation:

$E\left[{\cos }^2\theta \right]=\underset{-\pi }{\overset{+\pi }{\int }}\frac{1}{2\pi }Co{s}^2\theta d\theta$

$=\frac{1}{2\pi }\underset{-\pi }{\overset{+\pi }{\int }}\left(\frac{1+\cos 2\theta }{2}\right)d\theta$

$\Rightarrow \frac{1}{4\pi }\left[\underset{-\pi }{\overset{+\pi }{\int }}d\theta +\underset{-\pi }{\overset{+\pi }{\int }}cos2\theta d\theta \right]$

$\Rightarrow \frac{1}{4\pi }\left(2\pi \right)=\frac{1}{2}$

$E\left[{\left({\cos }^2\theta \right)}^2\right]=\frac{1}{2\pi }\underset{-\pi }{\overset{+\pi }{\int }}{\left(Co{s}^2\theta \right)}^{2}d\theta$

$=\frac{1}{2\pi }\underset{-\pi }{\overset{+\pi }{\int }}{\left(\frac{1+\cos 2\theta }{2}\right)}^{2}d\theta$

$=\frac{1}{8\pi }\underset{-\pi }{\overset{+\pi }{\int }}{\left(Co{s}^2\theta \right)}^{2}d\theta$

= 3/8

Variance = $\left(\frac{3}{8}\right)-{\left(\frac{1}{2}\right)}^2$

$=\frac{3}{8}-\frac{1}{4}$

$=\frac{1}{8}$