Respuesta :
Answer: a) P(x≤1) = 0.0625
b) P(0.5≤x≤1) = 0.0468
c) P(x>1.5) = 0.8593
d) Median (m) = 3.4641
e) f(x) = 0 x < 0
x/8 0≤x≤4
0 x≥4
Step-by-step explanation:
a) To calculate the probability of F(x) with x being less or equal 1 is:
P(x≤1) = F(1)
F(1) = [tex]\frac{1^{2}}{16}[/tex]
P(x≤1) = 0.0625
b) To calculate probability of F(x) with x being between 0.5 and 1:
P(0.5≤x≤1) = F(x≤1) - F(x≤0.5) = F(1) - F(0.5)
F(1) = 0.0625
F(0.5) = [tex]\frac{0.5^{2}}{16}[/tex] = 0.0156
P(0.5≤x≤1) = 0.0625 - 0.0156
P(0.5≤x≤1) = 0.0468
c) P(x>1.5) = 1 - P(x≤1.5) = 1 - F(1.5)
F(1.5) = 0.1406
P(x>1.5) = 1 - 0.1406
P(x>1.5) = 0.8593
d) Median is a point in the graph that divides it in half, so to determine the point, here called m:
[tex]\int\limits^m_0 {\frac{x^{2}}{16} } \, dx = \frac{1}{2}[/tex]
[tex]\frac{1}{16}\int\limits^m_0 {x^{2}} \, dx = \frac{1}{2}[/tex]
[tex]\int\limits^m_0 {x^{2}} \, dx = 8[/tex]
[tex]\frac{x^{3}}{3} = 8[/tex]
[tex]\frac{m^{3}}{3} - 0 = 8[/tex]
m³ = 24
m = 3.4641
The median checkout duration is 3.4641 hours.
e) Density function of a cumulative distribution function (cdf) as well as in a continuous random variable is the first derivative of a function. Then, for this function it is:
f(x) = F'(x)
f(x) = 0 x<0
[tex]\frac{x}{8}[/tex] 0≤x≤4
0 x≥4
f) E(X) = [tex]\int\limits^4_0 {x}f(x) \, dx[/tex]
= [tex]\int\limits^4_0 {x}.\frac{x}{8} \, dx[/tex]
= [tex]\int\limits^4_0 {\frac{x^{2}}{8} \, dx[/tex]
= [tex]\frac{x^{3}}{24}[/tex]
= [tex]\frac{4^{3}}{24} - \frac{0^{3}}{24}[/tex]
E(X) = 2.6667
