Continuous Uniform Distribution MCQ Quiz - Objective Question with Answer for Continuous Uniform Distribution - Download Free PDF

Concept:

For a uniform distribution defined over the interval $[a,b]$.The formulas for the mean and variance are:

• Mean: $\mu =\frac{a+b}{2}$
• Variance: ${\sigma }^2=\frac{(b-a{)}^2}{12}$

Given:

a = -4 & b =1

Calculation:

Mean:

$\mu =\frac{-4+1}{2}=\frac{-3}{2}=-1.5$

Variance:

${\sigma }^2=\frac{(1-(-4){)}^2}{12}=\frac{(5{)}^2}{12}=\frac{25}{12}$

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

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

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

Concept:

Uniform Distribution:

If x is continuous Random variable defined in [a, b], such that its p.d.f is given as:

$f\left(x\right)=\{\begin{array}{c}\frac{1}{b-a};a\le x\le b\\ 0;otherwise\end{array}$

Then x is called uniform Random variable.

 Continuous probability Distribution:

Let x is Continuous Random Variable, having p.d.f.  f(x), then we have following results:

1) f(x) = Prob at x

2) f(x) ≥ 0

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

4) $P\left(a\le x\le b\right)=\underset{a}{\overset{b}{\int }}f\left(x\right)$

5) $E\left(x\right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}xf\left(a\right)dx$ 

Calculation:

Let x be the length of the shorter piece of pencil. Hence it varies between $\left[0-\frac{1}{2}\right].$

By using uniform distribution, p.d.f of the given condition:

$f\left(x\right)=\{\begin{array}{c}\frac{1}{b-a};a<x\le b\\ 0,otherwise\end{array}$

$=\{\begin{array}{c}\frac{1}{\frac{1}{2}-0}=2;0<x<\frac{1}{2}\\ 0\end{array}$

From the definition of continuous probability distribution,

$E\left(x\right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}xf\left(x\right)dx=\underset{0}{\overset{\frac{1}{2}}{\int }}x\cdot 2dx=\frac{1}{4}$

Concept:

Uniform distribution:

$f\left(x\right)=\{\begin{array}{c}\frac{1}{b-a}fora<x<b\\ 0otherwise\end{array}$

Expectation $E\left(x\right)=\frac{a+b}{2}$ 

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

Calculation:

The area under the curve must be 1,

$\Rightarrow \frac{1}{4}\left(k-1\right)=1$ 

⇒ k = 5

Variance $=\frac{{\left(b-a\right)}^2}{12}=\frac{{\left(5-1\right)}^2}{12}=\frac{16}{12}=\frac{4}{3}$ 

Uniformly distributed means values placed between -2 to 3 on X – axis.

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

But we consider only between -2 and 3.

Case that will occur here is P(X) when X < 4

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

here, b= 3, a= -2

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

Concept:

In such questions, show various regions represented by equations on the graph.

Calculation:

Given that

0 ≤ x ≤ 10

0 ≤ y ≤ 20

p {x + y ≥ 20} = ?

Required probability = Area of right angled triangle ABC/Area of rectangular region OABD

$p=\frac{\frac{1}{2}\times 10\times 10}{10\times 20}=\frac{1}{4}=0.25$

Since Y is uniformly distributed in (1, 6)

$f\left(Y\right)=\frac{1}{b-a},1<Y<6$

Otherwise f(Y) = 0

3x² + 6xY + 3Y + 6

Comparing with

ax² + bx + c = 0

∴ a = 3, b = 6Y, c = 3Y + 6

Also, it has only real roots

b² - 4ac ≥ 0

(6Y)² - 4(3) (3Y + 6) ≥ 0

Y² - Y - 2 ≥ 0

(Y - 2) (Y + 1) ≥ 0

Product of two number (assume a and b) is ≥ 0 if and only if

1. a > 0 and b > 0

2. a < 0 and b < 0

3. a = 0 or b = 0

Condition 1:

Y - 2 > 0 and Y + 1 > 0

Y > 2 and Y > -1

∴ Y > 2

Condition 2:

Y - 2 < 0 and Y + 1 < 0

Y < 2 and Y < -1

∴ Y < -1

Condition 3:

Y - 2 = 0 or Y + 1 = 0

∴ Y = 2 or Y = -1

Value of Y satisfying the condition of real roots falls in the range: [2, 6)

area = Base × height

$area=\left(2-6\right)\times \frac{1}{5}$

∴ area = 0.8

Therefore, the probability that the polynomial 3x² + 6xY + 3Y + 6 has only real roots is 0.8

Tips and Tricks:

From the diagram:

If diagram is divided into 5 different parts (1, 6), then range which satisfy condition that the equation has only has real roots occupy the 4 parts [2, 6) of it.

Concept:

The probability density function is always positive f(x) ≥ 0, and it follows the below condition.

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

Calculation:

Given:

$f\left(x\right)=\{\begin{array}{cc}0.25& if1\le x\le 5\\ 0& otherwise\end{array}$

The probability density function of a random variable X is,

$P\left(x\le 4\right)=\underset{-\mathrm{\infty }}{\overset{4}{\int }}f\left(x\right)dx$

$=\underset{-\mathrm{\infty }}{\overset{1}{\int }}\left(0\right)dx+\underset{1}{\overset{4}{\int }}\left(0.25\right)dx$

$=\frac{1}{4}{\left(x\right)}_1^4=\frac{1}{4}\left(4-1\right)$

$\therefore P(x\le 4)=\frac{3}{4}$

Concept:

For a uniform distribution defined over the interval $[a,b]$.The formulas for the mean and variance are:

• Mean: $\mu =\frac{a+b}{2}$
• Variance: ${\sigma }^2=\frac{(b-a{)}^2}{12}$

Given:

a = -4 & b =1

Calculation:

Mean:

$\mu =\frac{-4+1}{2}=\frac{-3}{2}=-1.5$

Variance:

${\sigma }^2=\frac{(1-(-4){)}^2}{12}=\frac{(5{)}^2}{12}=\frac{25}{12}$

Concept:

In such questions, show various regions represented by equations on the graph.

Calculation:

Given that

0 ≤ x ≤ 10

0 ≤ y ≤ 20

p {x + y ≥ 20} = ?

Required probability = Area of right angled triangle ABC/Area of rectangular region OABD

$p=\frac{\frac{1}{2}\times 10\times 10}{10\times 20}=\frac{1}{4}=0.25$

Since Y is uniformly distributed in (1, 6)

$f\left(Y\right)=\frac{1}{b-a},1<Y<6$

Otherwise f(Y) = 0

3x² + 6xY + 3Y + 6

Comparing with

ax² + bx + c = 0

∴ a = 3, b = 6Y, c = 3Y + 6

Also, it has only real roots

b² - 4ac ≥ 0

(6Y)² - 4(3) (3Y + 6) ≥ 0

Y² - Y - 2 ≥ 0

(Y - 2) (Y + 1) ≥ 0

Product of two number (assume a and b) is ≥ 0 if and only if

1. a > 0 and b > 0

2. a < 0 and b < 0

3. a = 0 or b = 0

Condition 1:

Y - 2 > 0 and Y + 1 > 0

Y > 2 and Y > -1

∴ Y > 2

Condition 2:

Y - 2 < 0 and Y + 1 < 0

Y < 2 and Y < -1

∴ Y < -1

Condition 3:

Y - 2 = 0 or Y + 1 = 0

∴ Y = 2 or Y = -1

Value of Y satisfying the condition of real roots falls in the range: [2, 6)

area = Base × height

$area=\left(2-6\right)\times \frac{1}{5}$

∴ area = 0.8

Therefore, the probability that the polynomial 3x² + 6xY + 3Y + 6 has only real roots is 0.8

Tips and Tricks:

From the diagram:

If diagram is divided into 5 different parts (1, 6), then range which satisfy condition that the equation has only has real roots occupy the 4 parts [2, 6) of it.

Concept:

The probability density function of a uniform random variable on the interval (a,b) is given by:

\(f(x)=\left\{\begin{matrix} \frac{1}{b-a} &if a

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

$Variance{\sigma }^2=\frac{(b-a{)}^2}{12}$

$StandardDeviation\sigma =\sqrt{Variance}$

Calculation:

Given:

Mean = 1, Varaince = 4/3, 

For uniform distribution:

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

$1=\frac{a+b}{2}$

a + b = 2 ...... (1)

$Variance{\sigma }^2=\frac{(b-a{)}^2}{12}$

$\frac{4}{3}=\frac{(b-a{)}^2}{12}$

b - a = 4 ......(2)

Solving (1) and (2) we get

b = 3, a = -1

Therefore, the uniform distribution will be:

\(f(x)=\left\{\begin{matrix} \frac{1}{4} &if -1

${p(x<0)=\underset{-1}{\overset{0}{\int }}f\left(x\right)dx=\underset{-1}{\overset{0}{\int }}\frac{1}{4}dx=\frac{1}{4}x|}_{-1}^0=\frac{1}{4}$

Concept:

The probability density function is always positive f(x) ≥ 0, and it follows the below condition.

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

Calculation:

Given:

$f\left(x\right)=\{\begin{array}{cc}0.25& if1\le x\le 5\\ 0& otherwise\end{array}$

The probability density function of a random variable X is,

$P\left(x\le 4\right)=\underset{-\mathrm{\infty }}{\overset{4}{\int }}f\left(x\right)dx$

$=\underset{-\mathrm{\infty }}{\overset{1}{\int }}\left(0\right)dx+\underset{1}{\overset{4}{\int }}\left(0.25\right)dx$

$=\frac{1}{4}{\left(x\right)}_1^4=\frac{1}{4}\left(4-1\right)$

$\therefore P(x\le 4)=\frac{3}{4}$

Concept:

Uniform Distribution:

If x is continuous Random variable defined in [a, b], such that its p.d.f is given as:

$f\left(x\right)=\{\begin{array}{c}\frac{1}{b-a};a\le x\le b\\ 0;otherwise\end{array}$

Then x is called uniform Random variable.

 Continuous probability Distribution:

Let x is Continuous Random Variable, having p.d.f.  f(x), then we have following results:

1) f(x) = Prob at x

2) f(x) ≥ 0

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

4) $P\left(a\le x\le b\right)=\underset{a}{\overset{b}{\int }}f\left(x\right)$

5) $E\left(x\right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}xf\left(a\right)dx$ 

Calculation:

Let x be the length of the shorter piece of pencil. Hence it varies between $\left[0-\frac{1}{2}\right].$

By using uniform distribution, p.d.f of the given condition:

$f\left(x\right)=\{\begin{array}{c}\frac{1}{b-a};a<x\le b\\ 0,otherwise\end{array}$

$=\{\begin{array}{c}\frac{1}{\frac{1}{2}-0}=2;0<x<\frac{1}{2}\\ 0\end{array}$

From the definition of continuous probability distribution,

$E\left(x\right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}xf\left(x\right)dx=\underset{0}{\overset{\frac{1}{2}}{\int }}x\cdot 2dx=\frac{1}{4}$

Concept:

Uniform distribution:

$f\left(x\right)=\{\begin{array}{c}\frac{1}{b-a}fora<x<b\\ 0otherwise\end{array}$

Expectation $E\left(x\right)=\frac{a+b}{2}$ 

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

Calculation:

The area under the curve must be 1,

$\Rightarrow \frac{1}{4}\left(k-1\right)=1$ 

⇒ k = 5

Variance $=\frac{{\left(b-a\right)}^2}{12}=\frac{{\left(5-1\right)}^2}{12}=\frac{16}{12}=\frac{4}{3}$