The probability density function (p. d.f.) of a continuous random variable XX is defined to be: f(x)={x6+k for 0

The probability density function (p.d.f.) of a continuous random variable X is defined to be:

f(x)= x/6+k for 0<x<2.50 otherwise,

for some constant k.

For these problems, please ensure your answers are accurate to within 3 decimals.

Part a) Find the value of k that makes the above function a proper p.d.f.

Part b) Hence find P(0.5<X<1).

a. k = 0.192

b. P(0.5<X<1) = 1

Step-by-step explanation:

Given

f(x) = x/6 + k for 0 x < 2.50

To find the value of k that makes the above function a proper p.d.f.

∫ f(x) dx must be equal to 1

∫ f(x) dx = 1

Substitute f(x) = x/6 + k

∫ x/6 + k dx {0,2.50}

Integrate with respect to x

x²/(6*2) + kx {0,2.50} = 1

x²/12 + kx {0,2.50} = 1

(2.50²/12 + 2.50k) - (0²/12 - 0*k) = 1

6.25/12 + 2.50k = 1

Collect like terms

2.50k = 1 - 6.25/12

2.50k = (12 - 6.25)/12

2.50k = 5.75/12

Divide through by 2.50

2.50k/2.50 = 5.75/12 * 1/2.50

k = 2.3/12

k = 23/120

k = 0.192 --- Approx to 3 decimal places

So f(x) = x/6 + 23/120 for 0 x < 2.50

b.

Find P(0.5<X<1)

Given that f(x) = x/6 + 23/120 for 0 x < 2.50

P(0.5<X<1) = ∫x/6 + 23/120 {0,2.50}

P(0.5<X<1) = x²/12+ 23x/120 {0,2.50}

P(0.5<X<1) = (2.5²/12+ 23*2.5/120) - (0²/12+ 23*0/120)

P(0.5<X<1) = (2.5²/12+ 23*2.5/120) - (0)

P(0.5<X<1) = (6.25/12+ 57.5/120) - (0)

P(0.5<X<1) = (6.25/12+ 5.75/12)

P(0.5<X<1) = (6.25 + 5.75)/12

P(0.5<X<1) = 12/12

P(0.5<X<1) = 1

the answer is here: : )

:

for this pair of functions g(x)=x-2 and f(x)=|x| is (g o f)(a)= |a| -2

step-by-step explanation:

g(x)=x-2

f(x)=|x|

(g o f)(a)=?

(g o f)(x)=g(f(x))

(g o f)(x)=g(|x|)

(g o f)(x)=|x|-2

(g o f)(a)=|a|-2