Question

Water pump failure data have been analyzed, and it is found through a statistical goodness-of-fit test that the pump failure times can be described adequately by a triangular distribution with a = 849 hours, b = 8705 hours, and c = 5901 hours. Its probability density function is illustrated in the following figure. Figure 1 PDF Function Plot Find the average, the standard deviation, the median, and the 80th percentile of the life of such pumps. 1) The average of the life of such pumps is: E(t)= 2) The standard deviation of the life of such pumps is: Std(t)= 3) The median of the life of such pumps is: T50= 4) The 80th percentile of the life of such pumps is: T80=

Answer 1

1)

From Triangular distribution,

E(t) = (a + b + c)/3 = (849 + 8705 + 5901)/3 = 5151.667 hours

2)

Var(t) = (a² + b² + c² - ab - bc - ca)/18

= (849² + 8705² + 5901² - 849 * 8705 - 8705 * 5901 - 5901 * 849)/18

= 2641718

Std(t) = = 1625.336

2641718

3)

(a+b)/2 = (849 + 8705)/2 = 4777

Since, c $\ge$ (a+b)/2,

Median, T50 = a +

(6 - a)(c-a)/2

= 849 +

(8705 – 849) (5901 - 849)/2

= 5303.689

4)

CDF of Triangular distribution is,

F(x) = (x-a)² / (b-a) (c-a) for a < x $\le$ c

0.8 = (x - 849)² / (8705 - 849) (5901 - 849)

(x - 849)² / 39688512 = 0.8

x - 849 =

V0.8 * 39688512

x - 849 = 5634.786

x = 849 + 5634.786 = 6483.786

T80 = 6483.786