Chapter 5: Q. 5.4 (page 176)

The random variable $X$ has the probability density function
$f (x) = \{\begin{cases} a x + b x 2 & 0 < x < 1 \\ 0 & o t h e r w i s e \end{cases}$
If $E [X] = 6$ , find
(a) $P {X < \frac{1}{2}}$ and
(b) $V a r (X)$ .

Short Answer

Expert verified

(a) The value of $P {X < \frac{1}{2}}$ is $\frac{7}{20}$

(b) The value of $V a r (X)$ is $0.06$

Step by step solution

Find the value of P{X<12} (part a)

First and foremost, the density function must satisfy the following conditions:

$1 = \int_{ℝ} f (x) d x = \int_{0}^{1} (a x + b x^{2}) d x = a \times \frac{x^{2}}{2} + {b \times \frac{x^{3}}{3}|}_{0}^{1} = \frac{a}{2} + \frac{b}{3}$

We can deduce from the fact that $E (X) = 0.6$ ,

$0.6 = E (X) = \int_{ℝ} x f (x) d x = \int_{0}^{1} (a x^{2} + b x^{3}) d x = a \times \frac{x^{3}}{3} + {b \times \frac{x^{4}}{4}|}_{0}^{1} = \frac{a}{3} + \frac{b}{4}$

As a result, we have two unknowns and two linear equations to solve.

$\{\begin{cases} 3 a + 2 b & = 6 \\ 20 a + 15 b & = 36 \end{cases}$

$a = \frac{18}{5}$ and $b = - \frac{12}{5}$ are the results of this system.

$P (X < \frac{1}{2}) = \int_{0}^{1 / 2} \frac{18}{5} x - \frac{12}{5} x^{2} d x = \frac{9}{5} x^{2} - {\frac{4}{5} x^{3}|}_{0}^{1 / 2} = \frac{7}{20}$

Find Var(X) (part b)

Let's figure out the second moment in order to calculate the variance.

$E (X^{2}) = \int_{ℝ} x^{2} f (x) d x = \int_{0}^{1} \frac{18}{5} x^{3} - \frac{12}{5} x^{4} d x = \frac{18}{5} \frac{x^{4}}{4} - {\frac{12}{5} \frac{x^{5}}{5}|}_{0}^{1} = \frac{21}{50}$

As a result, we may say that the variance is equal to

$Var (X) = E (X^{2}) - E (X)^{2} = \frac{21}{50} - 0 . 6^{2} = 0.06$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

The random variable $X$ has the probability density function
$f (x) = \{\begin{cases} a x + b x 2 & 0 < x < 1 \\ 0 & o t h e r w i s e \end{cases}$
If $E [X] = 6$ , find
(a) $P {X < \frac{1}{2}}$ and
(b) $V a r (X)$ .

Short Answer

Step by step solution

Find the value of P{X<12} (part a)

Find Var(X) (part b)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Geometry

Statistics

Discrete Mathematics

Logic and Functions

Decision Maths

Study anywhere. Anytime. Across all devices.

The random variable Xhas the probability density functionf(x)=ax+bx20<x<10otherwiseIf E[X]=6, find(a) P{X<12}and(b) Var(X).

Short Answer

Step by step solution

Find the value of P{X<12} (part a)

Find Var(X) (part b)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Geometry

Statistics

Discrete Mathematics

Logic and Functions

Decision Maths

Study anywhere. Anytime. Across all devices.

The random variable $X$ has the probability density function
$f (x) = \{\begin{cases} a x + b x 2 & 0 < x < 1 \\ 0 & o t h e r w i s e \end{cases}$
If $E [X] = 6$ , find
(a) $P {X < \frac{1}{2}}$ and
(b) $V a r (X)$ .