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)*(a^{2}+b^{2}+m^{2}-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**

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