Q. 5.14 Suppose that the cumulative dist... [FREE SOLUTION]

91影视

A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 5: Q. 5.14 (page 218)

Suppose that the cumulative distribution function of the random variable $X$ is given by
$F (x) = 1 鈭� e^{鈭� x^{2}} x > 0$
Evaluate $(a) P [X > 2]; (b) P [1 < X < 3)$ ; (c) the hazard rate function of $F_{i} (d) E [X]; (e) Var 鈦� (X)$ .
Hint: For parts $(d)$ and $(e)$ , you might want to make use of the results of Theoretical Exercise $5.5$ .

Short Answer

Expert verified

(a) The cumulative distribution function of $P (X > 2) = e^{鈭� 4}$

(b) The cumulative distribution function of random variables is $P (1 < X < 3) = e^{鈭� 1} 鈭� e^{鈭� 9}$

(d) $E X = \frac{\sqrt{蟺}}{2}$

(e) $Var 鈦� (X) = \frac{3 蟺}{4}$

Step by step solution

Step :1 The cumulative distribution of the function P[X>2] (part a)

The cumulative distribution of the function $P [X > 2]$

$P (X > 2) = 1 鈭� P (X 鈮� 2) = 1 鈭� F_{X} (2) = 1 鈭� (1 鈭� e^{鈭� 2^{2}}) P (X > 2) = e^{鈭� 4}$

Step :2 The cumulative distribution of the function P[1<X<3) (part b)

$P (1 < X < 3) = P (X < 3) 鈭� P (X < 1) = F_{X} (3) 鈭� F_{X} (1) = e^{鈭� 1} 鈭� e^{鈭� 9} P (1 < X < 3) = e^{- 1} e^{- 9}$

Step :3 The hazard rate of function (part c)

We have that,

$f_{X} (x) = \frac{d}{d x} F_{X} (x) = \frac{d}{d x} (1 鈭� e^{鈭� x^{2}}) = 2 x e^{鈭� x^{2}} F_{X} (X) = 2 x e^{- x^{2}}$

Here the hazard function

$渭_{X} (x) = \frac{f_{X} (x)}{1 鈭� F_{X} (x)} = \frac{2 x e^{鈭� x^{2}}}{1 鈭� (1 鈭� e^{鈭� x^{2}})} 渭_{X} (X) = 2 x$

Step :4 Value of E[X] (part d)

$E X = {鈭�}_{鈩�} 鈥� x f_{X} (x) d x = {鈭�}_{0}^{鈭�} 鈥� x 鈰� 2 x e^{鈭� x^{2}} d x = {鈭�}_{0}^{鈭�} 鈥� 2 x^{2} e^{鈭� x^{2}} d x$

$u = x and d v = 2 x e^{鈭� x^{2}} d x . Thus d u = d x and v = 鈭� e^{鈭� x^{2}}$

${鈭�}_{0}^{鈭�} 鈥� 2 x^{2} e^{鈭� x^{2}} d x = 鈭� {x 鈰� e^{鈭� x^{2}}|}_{0}^{鈭�} + {鈭�}_{0}^{鈭�} 鈥� e^{鈭� x^{2}} d x = \frac{\sqrt{蟺}}{2}$

Step : 5 variable Var⁡(X)

In order to calculate the variance, let's find the second moment. We ve got

$E (X^{2}) = {鈭�}_{0}^{鈭�} 鈥� x^{2} f_{X} (x) d x = {鈭�}_{0}^{鈭�} 鈥� 2 x^{3} e^{鈭� x^{2}} d x$

$u = x^{2} and d v = 2 x e^{鈭� x^{2}} d x . Thus d u = 2 x d x and v = 鈭� e^{鈭� x^{2}}$

${鈭�}_{0}^{鈭�} 鈥� 2 x^{3} e^{鈭� x^{2}} d x = 鈭� {x^{2} 鈰� e^{鈭� x^{2}}|}_{0}^{鈭�} + {鈭�}_{0}^{鈭�} 鈥� 2 x e^{鈭� x^{2}} d x = {e^{鈭� x^{2}}|}_{0}^{鈭�} = 1$

Hence, the variance is adequate to

$Var 鈦� (X) = E (X^{2}) 鈭� E (X)^{2} = 1 鈭� \frac{蟺}{4} = \frac{3 蟺}{4}$

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Short Answer

Step by step solution

Step :1 The cumulative distribution of the function P[X>2] (part a)

Step :2 The cumulative distribution of the function P[1<X<3) (part b)

Step :3 The hazard rate of function (part c)

Step :4 Value of E[X] (part d)

Step : 5 variable Var⁡(X)

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91影视

Suppose that the cumulative distribution function of the random variable Xis given byF(x)=1鈭�e鈭�x2x>0Evaluate (a)P[X>2];(b)P[1<X<3); (c) the hazard rate function of Fi(d)E[X]; (e)Var鈦�(X).Hint: For parts (d)and (e), you might want to make use of the results of Theoretical Exercise 5.5.

Short Answer

Step by step solution

Step :1 The cumulative distribution of the function P[X>2] (part a)

Step :2 The cumulative distribution of the function P[1<X<3) (part b)

Step :3 The hazard rate of function (part c)

Step :4 Value of E[X] (part d)

Step : 5 variable Var⁡(X)

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

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Theoretical and Mathematical Physics

Geometry

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Discrete Mathematics

Calculus

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