Find the length of a line segment with endpoints of

A random variable X with a probability density function () = {^-x > 0

Sliva [168]

The solutions to the questions are

The probability that X is between 2 and 4 is 0.314
The probability that X exceeds 3 is 0.199
The expected value of X is 2
The variance of X is 2

<h3>Find the probability that X is between 2 and 4</h3>

The probability density function is given as:

f(x)= xe^ -x for x>0

The probability is represented as:

$P(x) = \int\limits^a_b {f(x) \, dx$

So, we have:

$P(2 < x < 4) = \int\limits^4_2 {xe^{-x} \, dx$

Using an integral calculator, we have:

$P(2 < x < 4) =-(x + 1)e^{-x} |\limits^4_2$

Expand the expression

$P(2 < x < 4) =-(4 + 1)e^{-4} +(2 + 1)e^{-2}$

Evaluate the expressions

P(2 < x < 4) =-0.092 +0.406

Evaluate the sum

P(2 < x < 4) = 0.314

Hence, the probability that X is between 2 and 4 is 0.314

<h3>Find the probability that the value of X exceeds 3</h3>

This is represented as:

$P(x > 3) = \int\limits^{\infty}_3 {xe^{-x} \, dx$

Using an integral calculator, we have:

$P(x > 3) =-(x + 1)e^{-x} |\limits^{\infty}_3$

Expand the expression

$P(x > 3) =-(\infty + 1)e^{-\infty}+(3+ 1)e^{-3}$

Evaluate the expressions

P(x > 3) =0 + 0.199

Evaluate the sum

P(x > 3) = 0.199

Hence, the probability that X exceeds 3 is 0.199

<h3>Find the expected value of X</h3>

This is calculated as:

$E(x) = \int\limits^a_b {x * f(x) \, dx$

So, we have:

$E(x) = \int\limits^{\infty}_0 {x * xe^{-x} \, dx$

This gives

$E(x) = \int\limits^{\infty}_0 {x^2e^{-x} \, dx$

Using an integral calculator, we have:

$E(x) = -(x^2+2x+2)e^{-x}|\limits^{\infty}_0$

Expand the expression

$E(x) = -(\infty^2+2(\infty)+2)e^{-\infty} +(0^2+2(0)+2)e^{0}$

Evaluate the expressions

E(x) = 0 + 2

Evaluate

E(x) = 2

Hence, the expected value of X is 2

<h3>Find the Variance of X</h3>

This is calculated as:

V(x) = E(x^2) - (E(x))^2

Where:

$E(x^2) = \int\limits^{\infty}_0 {x^2 * xe^{-x} \, dx$

This gives

$E(x^2) = \int\limits^{\infty}_0 {x^3e^{-x} \, dx$

Using an integral calculator, we have:

$E(x^2) = -(x^3+3x^2 +6x+6)e^{-x}|\limits^{\infty}_0$

Expand the expression

$E(x^2) = -((\infty)^3+3(\infty)^2 +6(\infty)+6)e^{-\infty} +((0)^3+3(0)^2 +6(0)+6)e^{0}$

Evaluate the expressions

E(x^2) = -0 + 6

This gives

E(x^2) = 6

Recall that:

V(x) = E(x^2) - (E(x))^2

So, we have:

V(x) = 6 - 2^2

Evaluate

V(x) = 2

Hence, the variance of X is 2

Read more about probability density function at:

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<u>Complete question</u>

A random variable X with a probability density function f(x)= xe^ -x for x>0\\ 0& else

a. Find the probability that X is between 2 and 4

b. Find the probability that the value of X exceeds 3

c. Find the expected value of X

d. Find the Variance of X

7 0

2 years ago

Which of the following shows the polynomial below written in descending order?

forsale [732]

To write the polynomial in descending order, arrange the terms from the highest to the lowest degree. Simply base on the exponents of the variable x. You have 3, 12, 1 and 7. In descending order, that would be: 12, 7, 3 and then 1. The answer is D.

4 0

3 years ago

Read 2 more answers

A ball is thrown vertically upward. After t seconds, its height h (in feet) is given by the function h(t) = 52t - 16t^2 . What i

Nata [24]

Answer: 42.25 feet

Step-by-step explanation:

We know that after "t" seconds, its height "h" in feet is given by this function:

$h(t) = 52t -16t^2$

The maximum height is the y-coordinate of the vertex of the parabola. Then, we can use the following formula to find the corresponding value of "t" (which is the x-coordinate of the vertex):

$x=t=\frac{-b}{2a}$

In this case:

$a=-16\\b=52$

Substituting values, we get :

$t=\frac{-52}{2(-16)}\\\\t=1.625$

Substituting this value into the function to find the maximum height the ball will reach, we get:

$h(1.625) = 52(1.625) -16(1.625)^2\\\\h(1.625) =42.25\ ft$

4 0

2 years ago

Read 2 more answers

You invest 1300 in a account that pays an interest rate of 4.75% what's the balance after 5 years ?

UNO [17]

Total = Principal * (1 + rate)^years
Total = 1,300 * (1.0475) ^ 5
Total = 1,300 * <span> <span> <span> 1.2611599139 </span> </span> </span>
Total = <span> <span> <span> 1,639.5</span></span></span>1

4 0

3 years ago

I am confused with this

Aleksandr-060686 [28]

Answer:

Answer is A) HL

Step-by-step explanation:

8 0

3 years ago

Find the length of a line segment with endpoints of –9 and 7.