Question

The total number of hours, measured in units of 100 hours, that a family runs a vacuum cleaner over a period of one year is a continuous random variable x that has the density function f(x) = x, 0 < x < 1, 2 x, 1 x < 2, 0, elsewhere. Find the probability that over a period of one year, a family runs their vacuum cleaner (a) less than 120 hours; (b) between 50 and 100 hours.

Answer 1

Answer:

0.68

0.375

Step-by-step explanation:

f(x) =  x   , 0 < x < 1

f(x) = 2 -x   1 ≤ x  < 2

f(x) = 0  for x elsewhere

The total number of hours, measured in units of 100 hours

less than 120 hours

x = 120/100 = 1.2

P (x < 1.2)  =  $\int\limits^1_0 {x} \, dx  + \int\limits^{1.2}_1 {2-x} \, dx$

=  ¹₀ [ x²/2 ]  +  ₁¹²[2x - x²/2]

= 1/2   +  (2 * 1.2  - 1.2²/2  - 2  + 1²/2)

= 0.5 + 2.4 - 0.72 - 2  + 0.5

= 0.68

between 50 and 100 hours.

x =50/100 = 0.5  & x = 100/100 = 1

P (0.5< x < 1)  =  $\int\limits^1_{0.5} {x} \, dx$

¹₀₅ [ x²/2 ]

= 1²/2 - 0.5²/2

= 0.5 - 0.125

= 0.375  = 3/8