Question

A variable follows the triangular distribution and has a lower limit of 200, an upper limit of 1600, and the most likely value of 750. Find the following:

a)Value of Standard Deviation

b)Probability the variable is less than 600?

c)Probability that variable is greater than 1000?

d)Probability the variable is between 800 and 1000?

Answer 1

for  triangular distribution

a=	200
b=	1600
m =	750

a)

V(X)=(1/18)*(a2+b2+m2-ab-am-bm)=			82916.667
standard deviation =σ =			287.953

b)

Probability the variable is less than 600

P(X<600 ) =(x-a)2/((b-a)*(m-a)) =(600-200)^2/((1600-200)*(750-200))=

0.2078

c)

for 1000 >750:

Probability that variable is greater than 1000 :

P(X>1000)=(b-x)2/((b-a)*(b-m))=(1600-1000)^2/((1600-200)*(1600-750))

=0.3025

d)

Probability the variable is between 800 and 1000 :

P(800 <X<1000) =P(X<1000)-P(X<800)

=(1600-800)^2/((1600-200)*(1600-750))-(1600-1000)^2/((1600-200)*(1600-750))

=0.2353