The probability distribution of the random variable X .

Question

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Answer 1

Question

Chapter 8.BMO, Problem 1BMO

To determine

The probability distribution of the random variable $X$ .

Expert Solution & Answer

Answer to Problem 1BMO

Solution:

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.05$	$0.1$	$0.25$	$0.3$	$0.2$	$0.1$

Explanation of Solution

Given:

The value taken on by a random variable $X$ and the frequency of their occurrence are shown in the following table.

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
Frequency of occurrence	$4$	$8$	$20$	$24$	$16$	$8$

Table $(1)$

Approach:

Let, $x1,x2,...xn$ , are the values of random variable $X$ , then the probability for $P(xi)$ is,

$P(X=xi)=xi∑xi$

Calculation:

Consider table $(1)$ , the sum of all the frequencies of occurrence is,

$∑xi=4+8+20+24+16+8=80$

Therefore, the sum of all frequencies of occurrence is $80$ .

Consider table $(1)$ , the probability of occurrence of $−3$ is,

$P(X=−3)=480=0.05$

The probability of occurrence of $−2$ is,

$P(X=−2)=880=0.1$

The probability of occurrence of $0$ is,

$P(X=0)=2080=0.25$

The probability of occurrence of $1$ is,

$P(X=1)=2480=0.3$

The probability of occurrence of $2$ is,

$P(X=2)=1680=0.2$

The probability of occurrence of $3$ is,

$P(X=3)=880=0.1$

Therefore, the probability distribution for random variable $X$ is,

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.05$	$0.1$	$0.25$	$0.3$	$0.2$	$0.1$

Table $(2)$

Conclusion:

Hence, the probability distribution for the random variable $X$ is shown in table $(2)$ .

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Answer 2

Question

Chapter 8.BMO, Problem 1BMO

To determine

The probability distribution of the random variable $X$ .

Expert Solution & Answer

Answer to Problem 1BMO

Solution:

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.05$	$0.1$	$0.25$	$0.3$	$0.2$	$0.1$

Explanation of Solution

Given:

The value taken on by a random variable $X$ and the frequency of their occurrence are shown in the following table.

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
Frequency of occurrence	$4$	$8$	$20$	$24$	$16$	$8$

Table $(1)$

Approach:

Let, $x1,x2,...xn$ , are the values of random variable $X$ , then the probability for $P(xi)$ is,

$P(X=xi)=xi∑xi$

Calculation:

Consider table $(1)$ , the sum of all the frequencies of occurrence is,

$∑xi=4+8+20+24+16+8=80$

Therefore, the sum of all frequencies of occurrence is $80$ .

Consider table $(1)$ , the probability of occurrence of $−3$ is,

$P(X=−3)=480=0.05$

The probability of occurrence of $−2$ is,

$P(X=−2)=880=0.1$

The probability of occurrence of $0$ is,

$P(X=0)=2080=0.25$

The probability of occurrence of $1$ is,

$P(X=1)=2480=0.3$

The probability of occurrence of $2$ is,

$P(X=2)=1680=0.2$

The probability of occurrence of $3$ is,

$P(X=3)=880=0.1$

Therefore, the probability distribution for random variable $X$ is,

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.05$	$0.1$	$0.25$	$0.3$	$0.2$	$0.1$

Table $(2)$

Conclusion:

Hence, the probability distribution for the random variable $X$ is shown in table $(2)$ .

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Answer 3

Question

Chapter 8.BMO, Problem 1BMO

To determine

The probability distribution of the random variable $X$ .

Expert Solution & Answer

Answer to Problem 1BMO

Solution:

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.05$	$0.1$	$0.25$	$0.3$	$0.2$	$0.1$

Explanation of Solution

Given:

The value taken on by a random variable $X$ and the frequency of their occurrence are shown in the following table.

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
Frequency of occurrence	$4$	$8$	$20$	$24$	$16$	$8$

Table $(1)$

Approach:

Let, $x1,x2,...xn$ , are the values of random variable $X$ , then the probability for $P(xi)$ is,

$P(X=xi)=xi∑xi$

Calculation:

Consider table $(1)$ , the sum of all the frequencies of occurrence is,

$∑xi=4+8+20+24+16+8=80$

Therefore, the sum of all frequencies of occurrence is $80$ .

Consider table $(1)$ , the probability of occurrence of $−3$ is,

$P(X=−3)=480=0.05$

The probability of occurrence of $−2$ is,

$P(X=−2)=880=0.1$

The probability of occurrence of $0$ is,

$P(X=0)=2080=0.25$

The probability of occurrence of $1$ is,

$P(X=1)=2480=0.3$

The probability of occurrence of $2$ is,

$P(X=2)=1680=0.2$

The probability of occurrence of $3$ is,

$P(X=3)=880=0.1$

Therefore, the probability distribution for random variable $X$ is,

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.05$	$0.1$	$0.25$	$0.3$	$0.2$	$0.1$

Table $(2)$

Conclusion:

Hence, the probability distribution for the random variable $X$ is shown in table $(2)$ .

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Answer 4

Question

Chapter 8.BMO, Problem 1BMO

To determine

The probability distribution of the random variable $X$ .

Expert Solution & Answer

Answer to Problem 1BMO

Solution:

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.05$	$0.1$	$0.25$	$0.3$	$0.2$	$0.1$

Explanation of Solution

Given:

The value taken on by a random variable $X$ and the frequency of their occurrence are shown in the following table.

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
Frequency of occurrence	$4$	$8$	$20$	$24$	$16$	$8$

Table $(1)$

Approach:

Let, $x1,x2,...xn$ , are the values of random variable $X$ , then the probability for $P(xi)$ is,

$P(X=xi)=xi∑xi$

Calculation:

Consider table $(1)$ , the sum of all the frequencies of occurrence is,

$∑xi=4+8+20+24+16+8=80$

Therefore, the sum of all frequencies of occurrence is $80$ .

Consider table $(1)$ , the probability of occurrence of $−3$ is,

$P(X=−3)=480=0.05$

The probability of occurrence of $−2$ is,

$P(X=−2)=880=0.1$

The probability of occurrence of $0$ is,

$P(X=0)=2080=0.25$

The probability of occurrence of $1$ is,

$P(X=1)=2480=0.3$

The probability of occurrence of $2$ is,

$P(X=2)=1680=0.2$

The probability of occurrence of $3$ is,

$P(X=3)=880=0.1$

Therefore, the probability distribution for random variable $X$ is,

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.05$	$0.1$	$0.25$	$0.3$	$0.2$	$0.1$

Table $(2)$

Conclusion:

Hence, the probability distribution for the random variable $X$ is shown in table $(2)$ .

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Answer 5

Chapter 8.BMO, Problem 1BMO

To determine

The probability distribution of the random variable $X$ .

Expert Solution & Answer

Answer to Problem 1BMO

Solution:

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.05$	$0.1$	$0.25$	$0.3$	$0.2$	$0.1$

Explanation of Solution

Given:

The value taken on by a random variable $X$ and the frequency of their occurrence are shown in the following table.

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
Frequency of occurrence	$4$	$8$	$20$	$24$	$16$	$8$

Table $(1)$

Approach:

Let, $x1,x2,...xn$ , are the values of random variable $X$ , then the probability for $P(xi)$ is,

$P(X=xi)=xi∑xi$

Calculation:

Consider table $(1)$ , the sum of all the frequencies of occurrence is,

$∑xi=4+8+20+24+16+8=80$

Therefore, the sum of all frequencies of occurrence is $80$ .

Consider table $(1)$ , the probability of occurrence of $−3$ is,

$P(X=−3)=480=0.05$

The probability of occurrence of $−2$ is,

$P(X=−2)=880=0.1$

The probability of occurrence of $0$ is,

$P(X=0)=2080=0.25$

The probability of occurrence of $1$ is,

$P(X=1)=2480=0.3$

The probability of occurrence of $2$ is,

$P(X=2)=1680=0.2$

The probability of occurrence of $3$ is,

$P(X=3)=880=0.1$

Therefore, the probability distribution for random variable $X$ is,

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.05$	$0.1$	$0.25$	$0.3$	$0.2$	$0.1$

Table $(2)$

Conclusion:

Hence, the probability distribution for the random variable $X$ is shown in table $(2)$ .

Answer 6

Chapter 8.BMO, Problem 1BMO

To determine

The probability distribution of the random variable $X$ .

Expert Solution & Answer

Answer to Problem 1BMO

Solution:

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.05$	$0.1$	$0.25$	$0.3$	$0.2$	$0.1$

Explanation of Solution

Given:

The value taken on by a random variable $X$ and the frequency of their occurrence are shown in the following table.

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
Frequency of occurrence	$4$	$8$	$20$	$24$	$16$	$8$

Table $(1)$

Approach:

Let, $x1,x2,...xn$ , are the values of random variable $X$ , then the probability for $P(xi)$ is,

$P(X=xi)=xi∑xi$

Calculation:

Consider table $(1)$ , the sum of all the frequencies of occurrence is,

$∑xi=4+8+20+24+16+8=80$

Therefore, the sum of all frequencies of occurrence is $80$ .

Consider table $(1)$ , the probability of occurrence of $−3$ is,

$P(X=−3)=480=0.05$

The probability of occurrence of $−2$ is,

$P(X=−2)=880=0.1$

The probability of occurrence of $0$ is,

$P(X=0)=2080=0.25$

The probability of occurrence of $1$ is,

$P(X=1)=2480=0.3$

The probability of occurrence of $2$ is,

$P(X=2)=1680=0.2$

The probability of occurrence of $3$ is,

$P(X=3)=880=0.1$

Therefore, the probability distribution for random variable $X$ is,

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.05$	$0.1$	$0.25$	$0.3$	$0.2$	$0.1$

Table $(2)$

Conclusion:

Hence, the probability distribution for the random variable $X$ is shown in table $(2)$ .

Answer 7

Chapter 8.BMO, Problem 1BMO

To determine

The probability distribution of the random variable $X$ .

Answer 8

Expert Solution & Answer

Answer to Problem 1BMO

Solution:

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.05$	$0.1$	$0.25$	$0.3$	$0.2$	$0.1$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

The value taken on by a random variable $X$ and the frequency of their occurrence are shown in the following table.

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
Frequency of occurrence	$4$	$8$	$20$	$24$	$16$	$8$

Table $(1)$

Approach:

Let, $x1,x2,...xn$ , are the values of random variable $X$ , then the probability for $P(xi)$ is,

$P(X=xi)=xi∑xi$

Calculation:

Consider table $(1)$ , the sum of all the frequencies of occurrence is,

$∑xi=4+8+20+24+16+8=80$

Therefore, the sum of all frequencies of occurrence is $80$ .

Consider table $(1)$ , the probability of occurrence of $−3$ is,

$P(X=−3)=480=0.05$

The probability of occurrence of $−2$ is,

$P(X=−2)=880=0.1$

The probability of occurrence of $0$ is,

$P(X=0)=2080=0.25$

The probability of occurrence of $1$ is,

$P(X=1)=2480=0.3$

The probability of occurrence of $2$ is,

$P(X=2)=1680=0.2$

The probability of occurrence of $3$ is,

$P(X=3)=880=0.1$

Therefore, the probability distribution for random variable $X$ is,

$x$	$−3$	$−2$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.05$	$0.1$	$0.25$	$0.3$	$0.2$	$0.1$

Table $(2)$

Conclusion:

Hence, the probability distribution for the random variable $X$ is shown in table $(2)$ .

Answer 12

Ch. 8.1 - What is a random variable? Give an example.Ch. 8.1 - Prob. 2CQ Ch. 8.1 - Prob. 3CQ Ch. 8.1 - Three balls are selected at random without...Ch. 8.1 - 2.A coin is tossed four times. Let the random...Ch. 8.1 - A die is rolled repeatedly until a 6 falls...Ch. 8.1 - Cards are selected one at a time without...Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

The probability distribution of the random variable X .

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The probability distribution of the random variable $X$ .

Answer to Problem 1BMO

Explanation of Solution

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The probability distribution of the random variable X .

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The probability distribution of the random variable X<m:math><m:mi>X</m:mi></m:math>.

Answer to Problem 1BMO

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The probability distribution of the random variable $X$ .