Respuesta :
Answer:
a.) f(x) = [tex]\frac{1}{30}[/tex] where 90 < x < 120
b.) [tex]\frac{2}{3}[/tex]
c.) [tex]\frac{2}{3}[/tex]
d.) [tex]\frac{1}{2}[/tex]
Step-by-step explanation:
Let
X be a uniform random variable that denotes the actual charging time of battery.
Given that, the actual recharging time required is uniformly distributed between 90 and 120 minutes.
⇒X ≈ ∪ ( 90, 120 )
a.)
Probability density function , f (x) = [tex]\frac{1}{120 - 90} = \frac{1}{30}[/tex] where 90 < x < 120
b.)
P(x < 110) = [tex]\int\limits^{110}_{90} {\frac{1}{30} } \, dx[/tex]
= [tex]\frac{1}{30}[x]\limits^{110}_{90} = \frac{1}{30} [ 110 - 90 ] = \frac{1}{30} [ 20] = \frac{2}{3}[/tex]
c.)
P(x > 100 ) = [tex]\int\limits^{120}_{100} {\frac{1}{30} } \, dx[/tex]
= [tex]\frac{1}{30}[x]\limits^{120}_{100} = \frac{1}{30} [ 120 - 100 ] = \frac{1}{30} [ 20] = \frac{2}{3}[/tex]
d.)
P(95 < x< 110) = [tex]\int\limits^{110}_{95} {\frac{1}{30} } \, dx[/tex]
= [tex]\frac{1}{30}[x]\limits^{110}_{95} = \frac{1}{30} [ 110 - 95 ] = \frac{1}{30} [ 15] = \frac{1}{2}[/tex]
