Probability Density Functions MCQ Quiz - Objective Question with Answer for Probability Density Functions - Download Free PDF

Given

f(x) = 1/√2πe^-x2/2dx

Formula

$E\left(y\right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}yf\left(x\right)dx$

Calculation

$E\left(y\right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}yf\left(x\right)dx=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\left(x+1\right)1/\surd 2\pi$(e^-x2/2dx

⇒ (1/√2π)${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\left(x+1\right)$e^-x2/2dx

⇒ E(y) = (1/√2π)$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}$x × e^-x2/2dx + (1/√2π)($\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}$e^-x2/2dx

⇒ Since, x × e^-x2/2dx is an odd function of k 

∴ (1/2√2π)($\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}$e^-x2/2dx = 0

⇒ E(y) = (1/2√2π)($\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}$e^-x2/2dx

⇒ (1/√2π) × √2π = 1

Using the standard integrals we have

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}$e^-x2/2dx  = √2π 

Formula

E(x/y = 1) = ∑x_j P(x = xy/y = 1

Calculation

According to the formula

⇒ 1 × (0.4/0.3)

⇒ 4/3

∴ The E(x|y = 1) is: 4/3

Given

$f\left(x\right)=\{\begin{array}{cc}3x^2,& o\le x\le 1\\ 0,& elsewhere\end{array}$

Calculation

P(X < a) = 1/a² = ${\int }_0^{a}f\left(x\right)dx=1/2$

⇒ $3{\int }_0^a$x²dx = 1/2

⇒ 3[x³/3]^a₀ = 1/2

⇒ 3 × (a³/3 - 0) = 1/2

⇒ a³ = 1/2

∴ The value of a such that P[x ≤ a] = P[x > a] is (1/2)^1/3

Given

f(x) = k.|x|, if x = - 2, 1, 3

         0 otherwise

Formula

variance = E(x²) - (E(x))²

Concept

Total probability = 1

Calculation

$\sum _{x=-2}f\left(x\right)+\sum _{x=1}f\left(x\right)+\sum _{x=3}f\left(x\right)=1$

⇒ $k\sum _{x=-2}\left|X\right|+k\sum _{x=1}\left|X\right|+k\sum _{x=3}\left|X\right|=1$

⇒ k|-2| + k|1| + k|3| = 1

⇒ 2k + k + 3k = 1

⇒ k = 1/6

We know that

Var x = E(x2) - (E(x))²

E(x) = ∑xf(x)

⇒ E(x) = $\sum _{x=-2}xf\left(x\right)+\sum _{x=1}xf\left(x\right)+\sum _{x=3}xf\left(x\right)$

⇒ E(x) = $1/6\sum _{x=-2}x\left|x\right|+1/6\sum _{x=1}x\left|x\right|+1/6\sum _{x=3}x\left|x\right|$

⇒ E(x) = 1/6[ -2 × 2 + 1 × 1 + 3 × 3]

⇒ E(x) = 1/6[6]

⇒ E(x) = 1

Now E(x²) = ∑x²f(x)

⇒ E(x2) = $\sum _{x=-2}$x²f(x) + $\sum _{x=1}$x²f(x) + $\sum _{x=3}$x²f(x)

⇒ E(x2) =1/6[ $\sum _{x=-2}$x²|x| + $\sum _{x=1}$x²|x| + $\sum _{x=3}$x²|x|

⇒ E(x2) = 1/6[(-2)² × 2 + 1² × 1 + 3² × 3]

⇒ E(x2) = 1/6(8 + 1 + 27) = 6

⇒ Variance = 6 - (1)²

∴ Variance is 5

Given

Observations = -2, 2, 5, 9, -9, 7, -7

Pdf = $f\left(x\right)=\{\begin{array}{cc}{e}^{-\left(X-\theta \right)},& x>\theta \\ 0& elsewhere\end{array}$

Formula

Likelihood function is given by

$L\left(x;\theta \right)=\sum _{i=1}^n$f(x_i; θ)

Calculation

L(x; θ) = $\sum _{i=1}^n$e^{-(x - θ)}

Taking log on both side 

⇒ Log L(x; θ) = $\sum _{i=1}^n$log($\sum _{i=1}^n$e^{-(x - θ)}

⇒ $\sum _{i=1}^n$(x_i - θ ])loge)

Log L is not differentiable w.r.t θ

According to maximum likelyhood principle

Log l has to be maximised

Log L is maximum if ∑(x_i - θ) is minimum

The mean deviation about a constant θ  = $\sum _{i=1}^n$(x_i - θ)

When x is odd than θ  = Median(x₁, x₂, ----x_n)

∴  Median of (-2, 2, 5, 9, -9, 7, -7) is 9

∴ Then the MLE of θ is 9

$P\left(X\le \frac{2}{3}|X>\frac{1}{3}\right)=\frac{P\left(X\le \frac{2}{3}\cap X>\frac{1}{3}\right)}{P\left(X>\frac{1}{3}\right)}$

$=\frac{P\left(\frac{1}{3}<X\le \frac{2}{3}\right)}{P\left(X>\frac{1}{3}\right)}$

$P\left(\frac{1}{3}<X\le \frac{2}{3}\right)=\underset{1/3}{\overset{2/3}{\int }}4x^{3}dx={\left[x^4\right]}_{\frac{1}{3}}^{\frac{2}{3}}=\frac{15}{81}=\frac{5}{27}$

$P\left(X>\frac{1}{3}\right)=\underset{1/3}{\overset{1}{\int }}4x^{3}dx$

$={\left[x^4\right]}_{\frac{1}{3}}^1$

$=1-\frac{1}{81}=\frac{80}{81}$

$P\left(X\le \frac{2}{3}|X>\frac{1}{3}\right)=\frac{\frac{5}{27}}{\frac{80}{81}}=\frac{5}{27}\times \frac{81}{80}=\frac{3}{16}$

Given

f(x) = { ke^-3x, x > o

          { 0, elsewhere

Concept used

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f\left(x\right)dx$ = $\underset{-\mathrm{\infty }}{\overset{0}{\int }}f\left(x\right)dx$ + $\underset{0}{\overset{\mathrm{\infty }}{\int }}f\left(x\right)dx$ = 1

Calculation

According to given part

⇒$\underset{-\mathrm{\infty }}{\overset{0}{\int }}f\left(x\right)dx$ = 0

⇒ 0 + $\underset{0}{\overset{\mathrm{\infty }}{\int }}f\left(x\right)dx$ = 0

⇒ ∫ke^-3xdx = 1

⇒ k[-e^-3x/3]

⇒ -k/3[e^-∞- e⁰] = 1

⇒ -k/3(0 – 1) = 1

⇒ k./3 = 1

∴ The value of k = 3 for PDF

Explanation:

x₁ & x₂: two independent exponentially distributed random variables with means 0.5 and 0.25.

A continuous random variable x is said to have an exponential (λ) distribution if it has probability density function.

$f_x\left(x/\lambda \right)=\{\begin{array}{cc}\lambda e^{-\lambda x}& forx>0\\ 0& forx\le 0\end{array}$

Where, λ > 0, is called the rate of distribution. The mean of the exponential (λ) distribution is calculated using integration by parts as:

$E\left[x\right]=\underset{0}{\overset{\mathrm{\infty }}{\int }}x\lambda .e^{-\lambda x}dx$

$=\lambda \left\{{\left[\frac{-x.e^{-\lambda x}}{\lambda }\right]}_0^{\mathrm{\infty }}+\frac{1}{\lambda }\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-\lambda x}.dx\right\}$

$=\lambda \left\{\lambda \left[0+0\right]+\frac{1}{\lambda }{\left[\frac{-e^{-\lambda x}}{\lambda }\right]}_0^{\mathrm{\infty }}\right\}$

$=\lambda \left\{0+\frac{1}{\lambda }\left[0+\frac{1}{\lambda }\right]\right\}$

$=\lambda .\frac{1}{{\lambda }^2}$

$E\left[x\right]=\frac{1}{\lambda }$

Let x₁, x₂ …..., x_n be independent random variables, with x_i having exponential (λ_i) distribution. Then the distribution of min (x₁, x₂ ……, x_n) is exponential (λ₁ + λ₂ + …. + λ_n)

Thus, mean of min $\left(x_1,x_2\right)=\frac{1}{{\lambda }_1+{\lambda }_2}$

${\lambda }_1=\frac{1}{E\left[x_1\right]}=\frac{1}{0.5}=2$

${\lambda }_2=\frac{1}{E\left[x_2\right]}=\frac{1}{0.25}=4$

Mean of min $\left(x_1,x_2\right)=\frac{1}{2+4}$

${\lambda }_2=\frac{1}{E\left[x_2\right]}=\frac{1}{0.25}=4$

Given

PMF(x) = 1/2^x, x = 1, 2, 3 -----

Formula

P(X > 4) = 1 – P(X ≤ 4)

Calculation

P(X > 4) = 1 – P(X ≤ 4)

⇒ 1 – [P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)]

⇒ 1 – [1/2¹ + ½²  + ½³ + ½⁴]

⇒ 1 – [1/2 + ¼ + 1/8 + 1/16]

⇒ 1 – [(8 + 4 + 2 + 1)/16]

⇒ 1 – [15/16]

⇒ (16 – 15)/16

∴ The value of P(X > 4) is 1/16

Probability mass function or  PMF is defined as the set of ordered pair [x, f(x)] if for each possible outcome x, f(x) satisfies the following condition

F(x) ≥ 0

∑ f(x) = 1

Concept:

If f(x) is probability density function for random variable x then,

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

Calculation:

We are given a probability density function

$f\left(x\right)=K\cdot \frac{1}{1+x^2}forx\in \left(-\mathrm{\infty },\mathrm{\infty }\right)$

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(x\right)dx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{K}{1+x^2}dx$

$=K{\left[{\tan }^{-1}x\right]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}=1$

$=K\left[\frac{\pi }{2}-\left(-\frac{\pi }{2}\right)\right]=1$

⇒ K(π) = 1

Given

F(x) = e^-x

Calculation

 For $Q_3={\displaystyle {\int }_0^{Q_3}{e}^{-x}dx={\displaystyle \frac{75}{100}}}$

⇒ ¾ = (-e^-x)^Q3₀

⇒ ¾ = (-1 – e^-Q3)

⇒ ¼ = e^-Q3

⇒Q₃ = In 4

For $Q_1={\displaystyle {\int }_0^{Q_1}{e}^{-x}dx={\displaystyle \frac{25}{100}}}$

⇒ (-e^-xdx)^Q1₀ = 1/4

⇒ (- 1 – e^-Q1) = ¼

⇒ Log Q₁ = ¾

Q₁ = In 4/3

⇒ Interquartile range = Q₃ – Q₁

⇒ In 4 – In 4/3

⇒ In  4 –(In 4 – In 3)

∴ The interquartile range is In 3

The Probability density function = ${\displaystyle {\int }_0^{\mathrm{\infty }}f(x)dx=1}$

Concept:

For a given probability density function f(x) to be valid, the integration of it over the complete range must equal 1.

Mathematically, it must satisfy:

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

Application:

Given f(x) = a + bx, 0 ≤ x ≤ 1

For the given density function to be valid, it must satisfy:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}a+bx=1$

$\underset{0}{\overset{1}{\int }}\left(a+bx\right)dx=1$

$\begin{array}{l}ax+{\frac{b{x}^2}{2}]}_0^1=1\\ a+\frac{b}{2}=1\end{array}$

To satisfy the above equation:

a = 0.5, b = 1

Expalantion

For the 5^th Decile = $$

Calculation

$$ = = ¾(2x²/2 – x³/3)₀^d5

⇒ 5/10 = ¾(2D₅²/2 – D₅³/3)

Putting the values of D₅, Put the value of D₅ in the equation, the value which satisfy the equation will become the answer of this question

Option 1 – (1)

⇒ ¾(2 × (1)²/2_– (1)³/3)

⇒ ¾(1 – 1/3

⇒ ¾(2/3) = ½ or 5/10

∴ Option 1 satisfy the equation hence 5^th decile point of X is 1                                                                                     

Concept: 

If f(x) is probability density function, then

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

2. Mean $=E\left(x\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x\cdot f\left(x\right)dx$

Calculation:

X has a probability density function f(x)

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}=1$

$\Rightarrow \underset{0}{\overset{\mathrm{\infty }}{\int }}\mathrm{k}{\mathrm{x}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{x}}\mathrm{d}\mathrm{x}=1$

We know that,

$\mathrm{\Gamma }\mathrm{m}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-\mathrm{x}}{\mathrm{x}}^{\mathrm{m}-1}\mathrm{d}\mathrm{x}$

$\Rightarrow \mathrm{k}\mathrm{\Gamma }\left(\mathrm{n}+1\right)=1$ → (1)

Given that mean = 3

$\Rightarrow \underset{0}{\overset{\mathrm{\infty }}{\int }}\mathrm{x}\mathrm{f}\left(x\right)\mathrm{d}\mathrm{x}=3$

$\Rightarrow \underset{0}{\overset{\mathrm{\infty }}{\int }}\mathrm{k}{\mathrm{x}}^{\mathrm{n}+1}{\mathrm{e}}^{-\mathrm{x}}\mathrm{d}\mathrm{x}=3$ 

$\Rightarrow \mathrm{k}\mathrm{\Gamma }\left(\mathrm{n}+2\right)=3$ → (2)

From equations (1) and (2)

$\Rightarrow \frac{k\mathrm{\Gamma }\left(n+1\right)}{k\mathrm{\Gamma }\left(n+2\right)}=\frac{1}{3}$

$\Rightarrow 3\mathrm{\Gamma }\left(n+1\right)=\mathrm{\Gamma }\left(n+2\right)$

We know that $\mathrm{\Gamma }n=\left(\mathrm{n}-1\right)!$

⇒ 3n! = (n + 1)!

⇒ 3 = n + 1

⇒ n = 2

From equation (1),

Calculation:

To find the first quartile of X, we need to find the value x such that the area under the probability density function f(x) to the left of x is 0.25.

We can begin by finding the cumulative distribution function (CDF) of X:

F(x) = ∫[0, x] f(t) dt

Since f(x) is a uniform distribution between 0 and 5, we have:

f(x) = 1/5 for 0 ≤ x ≤ 5

Therefore, the CDF of X is:

F(x) = ∫[0, x] (1/5) dt = x/5

To find the first quartile of X, we need to find the value x such that:

F(x) = x/5 = 0.25

Solving for x, we get:

x = 0.25 × 5 = 1.25

Therefore, the first quartile of X is 1.25.