Mathematics, 17.01.2020 21:31, savyblue1724707

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

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answered: allieeastridge

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!

answered: Guest

idk

step-by-step explanation:

idk

answered: Guest

solution:

we have to construct

∠m nt ≅ ∠p qr

the steps to draw this construction are

step 1: use a compass to draw an arc from point q which intersects the side pq at point a and the side qr at point b.

step 2: draw a segment nt and use the same width of the compass to draw an arc from point n which intersects the segment nt at a point x.

step 3:

step 4: join points n and y using a straightedge.

step 3:

option (d).maintaining the same width of the compass as ab, draw an arc from point x such that it intersects the arc drawn from n in a point y.

to draw congruent angles , 1. width of compasses while drawing two angles should be same i.e from point at which angle has to be drawn, 2.the distance between the two points where the compass intersects the two sides should be same while drawing any two congruent angles.