Answer to Question #246708 in Statistics and Probability for Akilesh

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

a)

x = 120/100 = 1.2

P (x < 1.2) = $\int^1_0xdx+\int^{1.2}_1(2-x)dx=$

= 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

b)

x =50/100 = 0.5 & x = 100/100 = 1

P (0.5< x < 1) =$\int^1_{0.5}xdx=$ ¹₀₅ [ x²/2 ]

= 1²/2 - 0.5²/2

= 0.5 - 0.125

= 0.375 = 3/8