Question

The following density function describes a random variable X. f(x) = 1 − (x /2) if 0

Answer 1

Answer:

a. Find the probability that X is greater than 1: _P(X>1) = 0.25

b. Find the probability that X is less than .5: _P(X<0.5)_

c. Find the probability that X is equal to 1.5:  P(X=1.5)= 0

Step-by-step explanation:

Hello!

The following density function describes a random variable X. f(x) = 1 − (x /2) if 0<x<2 a. Find the probability that X is greater than 1 ________ b. Find the probability that X is less than .5. _________ c. Find the probability that X is equal to 1.5.

First step to calculate the asked probabilities is to integrate the density function.

f(x) = 1 − (x /2) if 0<x<2

$\int\limits^2_0 {(1- (\frac{x}{2}))} \, dx$

$\int\limits^2_0 {1} \, dx - \frac{1}{2} \int\limits^2_0 {x} \, dx$

Now you resolve both integrals:

$\int\limits^2_0 {1} \, dx = x$

$\frac{1}{2} \int\limits^2_0 {x} \, dx = \frac{1}{2} * \frac{x^2}{2} = \frac{x^2}{4}$

$\int\limits^2_0 {(1- (\frac{x}{2}))} \, dx$ = $x-\frac{x^2}{4}$

The cummulative distribution is:

0 for x ≤ 0

$x-\frac{x^2}{4}$ for 0 < x < 2

1 for x ≥ 2

a. Find the probability that X is greater than 1.

P(X>1) = 1 - P(X ≤ 1)

"1" is included in the interval "0 < x < 2", to calculate the probability you have to replace it with $x-\frac{x^2}{4}$ and replace X with 1

1 - P(X ≤ 1) = 1 - ($1-\frac{1^2}{4}$)= 1 - 075= 0.25

b. Find the probability that X is less than 0.5.

"0.5" in included in the interval "0 < x < 2", to calculate the probability you have to replace it with $x-\frac{x^2}{4}$ and replace X with 0.5

P(X<0.5)= $0.5-\frac{0.5^2}{4}$= 0.4375

c. Find the probability that X is equal to 1.5.

"1.5" is included in the interval "0 < x < 2", to calculate the probability you have to replace it with $x-\frac{x^2}{4}$ and replace X with 1.5

This is a continuous variable, in this type of variable the cumulative probability of X=k (k= constant) is always cero.

You can prove it doing the following calculation:

$\int\limits^{1.5}_{1.5} {x-\frac{x^2}{4}} \, dx$= $1.5-\frac{1.5^2}{4}$ - ($1.5-\frac{1.5^2}{4}$) = 0

I hope it helps!

Answer 2

The correct answers are given below:

• The probability that X is greater than 1: _P(X>1) = 0.25
• The probability that X is less than .5: _P(X<0.5)_
• The probability that X is equal to 1.5:  P(X=1.5)= 0

Calculations and Parameters:

We would first calculate the asked probabilities and integrate the density function.

f(x) = 1 − (x /2) if 0<x<2

After the integration and resolution of the integrals,

The cumulative distribution is 0 for x ≤ 0

x-(x^2/4) for 0 < x < 2

1 for x ≥ 2

The probability that X is greater than 1 is

P(X>1) = 1 - P(X ≤ 1)

1 - P(X ≤ 1) = 1 - (1- 1^2/4)= 1 - 075

= 0.25

The probability that X is less than 0.5 is

P(X<0.5)= 0.5- (0.5^2/4)

= 0.4375

The probability that X is equal to 1.5 is =0

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