Continuous Distributions MCQ Quiz in मल्याळम - Objective Question with Answer for Continuous Distributions - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

The mean value of (μ) of the probability distribution of a variate X is commonly known as expectation and it is denoted by E[X].

If f(x) is the probability density function of the variate X, then

Discrete distribution: $E\left[X\right]=\sum _i{x}_{i}f\left(x_i\right)$

Continuous distribution: $E\left(X\right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}xf\left(x\right)dx$

Calculation:

$E\left(X^2\right)=\underset{0}{\overset{1}{\int }}{x}^2\left(3x^2\right)dx$

Explanation:

Normal/Gaussian/Bell distribution:

Probability distribution function (PDF) for a normal distribution is:

$PDF=f\left(x\right)=\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{1}{2}{\left(\frac{x-\mu }{\sigma }\right)}^2}$

where

x = normal random variable

μ = mean = mode = median

σ = standard deviation. σ² = variance

Probability is:

$P\left(a\le x\le b\right)=\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{1}{2}{\left(\frac{x-\mu }{\sigma }\right)}^2}dx\dots ..\left(1\right)$

Standard normal distribution:

In this case, $\frac{x-\mu }{\sigma }=z$

where z is a standard normal variable

Differentiating the above equation, we get

$\frac{dx}{\sigma }=dz\Rightarrow dx=\sigma dZ$

 Putting the values in equation (1)

$\underset{z_1}{\overset{z_2}{\int }}\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}{z}^2}\sigma dz=\underset{z_1}{\overset{z_2}{\int }}\frac{1}{\sqrt{2\pi }}{e}^{-\frac{1}{2}{z}^2}dz$

PDF is:

$f\left(z\right)=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{z^2}{2}}$

After rearranging, we get

$f\left(z\right)=\frac{1}{\sqrt{2\pi {\left(1\right)}^2}}{e}^{-\frac{1}{2}{\left(\frac{z-0}{1}\right)}^2}$

Mean = Mode = Median = 0

Standard deviation = 1

Variance = 1

Uniformly distributed in (-2, 3), on X-axis.

X ~ U(-2,3)

Comparing with X ~ U(a,b)

Sample space = {-2, -1, 0, 1, 2, 3}

Here, b= 3, a= -2

$Variance=\frac{1}{12}{\left(b-a\right)}^2$

$\mathrm{V}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}=\frac{1}{12}{\left(3-\left(-2\right)\right)}^2=\frac{25}{12}$

Mean of the uniform distributed variable is given as

$mean=\frac{a+b}{2}$

$\therefore mean=\frac{-2+3}{2}=\frac{1}{2}$

The given probability density function is

$f\left(x\right)=\{\begin{array}{c}0.2,for\left|x\right|\le 1\\ 0.1,for1<\left|x\right|\le 4\\ 0,otherwise\end{array}$

Now, $f\left(x\right)=\{\begin{array}{c}0.2,x=\left(-1,1\right)\\ 0.1,x=\left(-4,-1\right)\cup \left(1,4\right)\\ 0,otherwise\end{array}$

The graph will be,

The probability P(0.5 < X < 5)

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

$=\underset{0.5}{\overset{1}{\int }}0.2dx+\underset{1}{\overset{4}{\int }}0.1dx$

= 0.2 [1 – 0.5] + 0.1 [4 - 1]

P = 0.2 × 0.5 + 0.1 × 3

= 0.1 + 0.3 = 0.4 (Shaded area)

Probability distribution function (PDF) of an exponential distribution is

$f\left(x,\lambda \right)=\{\begin{array}{cc}\lambda e^{-\lambda x}& x\ge 0\\ 0& x<0\end{array}$

Cumulative distribution function of an exponential distribution is

$f\left(x,\lambda \right)=\{\begin{array}{cc}1-e^{-\lambda x}& x\ge 0\\ 0& x<0\end{array}$

where λ > 0 is the parameter of distribution. So only one parameter in exponential

distribution.

The normal (or Gaussion) distribution is a very common continuous probability distribution.

The probability density of normal distribution is

$\delta \left(\frac{x}{\mu },{\sigma }^2\right)=\frac{1}{\sqrt{2\pi \sigma }}{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}$

Concept:

For an exponential distribution, the probability density function will be given by 

f(p) = θ e^-θp

where mean = 1/θ and standard deviation = 1/θ  

Given mean = 0.5 ⇒ Variance = 0.25

Mean = E(P), Variance = σ² = E(P²) - (E(P))²

⇒ 0.25 = E(P²) - 0.5² ⇒ E(P2) = 0.25 + 0.25 = 0.50  

Explanation:

Normal/Gaussian/Bell distribution:

Probability distribution function (PDF) for a normal distribution is:

$PDF=f\left(x\right)=\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{1}{2}{\left(\frac{x-\mu }{\sigma }\right)}^2}$

Where,

x = normal random variable

μ = mean = mode = median

σ = standard deviation. σ2 = variance

For continuous distribution

Standard normal distribution:

In this case, $\frac{x-\mu }{\sigma }=z$

where z is a standard normal variable

Mean = Mode = Median = 0

Standard deviation = 1

Variance = 1

Explanation:

From the above discussion, we can say that,

Mean = 0 and Variance = 1

Hence, option 3 is correct.

Concept:

We know for probability density function

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(x\right)dx=1$

Calculation:

Given:

f(x) = λ (x – 1) × (2 – x) 1 ≤ x ≤ 2, 0 otherwise

${\int }_1^2\lambda \left(-x^2+3x-2\right)dx=1$

$\Rightarrow \lambda {\left[{\displaystyle \frac{-x^3}{3}}+3\times {\displaystyle \frac{x^2}{2}}-2x\right]}_1^2=1$

$\Rightarrow \lambda \left[{\displaystyle \frac{-14+27-12}{6}}\right]=1$

Concepts:

For probability density function:

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)dx=1$

where f(x) is probability density function

Calculation:

Given:

$f(x)=\{\begin{array}{cc}\frac{k{x}^2}{2}& -1\le x\le 1\\ 0& \mathrm{e}\mathrm{l}\mathrm{s}\mathrm{e}\mathrm{h}\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e},\end{array}$

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)dx=1$

${\int }_{-\mathrm{\infty }}^{-1}f(x)dx+{\int }_{-1}^{1}f(x)dx+{\int }_1^{\mathrm{\infty }}f(x)dx=1$

${\int }_{-\mathrm{\infty }}^{-1}0dx+{\int }_{-1}^1\frac{k{x}^2}{2}dx+{\int }_1^{\mathrm{\infty }}0dx=1$

$\frac{k{x}^3}{6}|\begin{array}{c}1\\ -1\end{array}$ = 1

∴ k = 3

The correct answer is (option 2) Mean.

Explanation:

Normal Distribution:

• Normal distribution is also known as the Gaussian distribution.
• It is a probability distribution that is symmetric about the mean,
  showing that data near the mean are more frequent in occurrence than data far from the mean.
• In graphical form, the normal distribution appears as a "bell curve".