Exponential Distribution

For lengths of rope, 4 faults occur every 5 metres.

How could we model the waiting interval (in this instance, length), X, until the first fault is observed?

• Binomial

• Geometric

• Exponential

• Poisson

• Yeah for sure

• nah

• what

For X~Exp(λ)

E(X) =

• λ

• 1/λ

• 1-λ

• 1+λ

• 1*λ

For X~Exp(λ)

Var(X) =

• λ²

• 1/λ²

• 1-λ²

• 1+λ²

• 1²*λ²

For lengths of rope, 4 faults occur every 5 metres.

X~Exp(λ)

Solve for λ (fraction or decimal)

(a)  

For lengths of rope, 4 faults occur every 5 metres.

X is the length until the first fault.

X~Exp(0.8)

Find the probability the first fault occurs between the 1st and 3rd metre (4dp).

(a)  

For lengths of rope, 4 faults occur every 5 metres.

X is the length until the first fault.

X~Exp(0.8)

Find the probability the first fault occurs between the 2nd and 4th metre (4dp).

(a)  

For lengths of rope, 4 faults occur every 5 metres.

X is the length until the first fault.

X~Exp(0.8)

Find the EXACT probability the first fault occurs between the 1st and 5th metre.

NOTE: let e be m

Mathematical Equivalence

ON

For lengths of rope, 4 faults occur every 5 metres.

X is the length until the first fault.

X~Exp(0.8)

Find the probability the first fault occurs within the first 2 metres (4dp).

(a)  

For lengths of rope, 4 faults occur every 5 metres.

X is the length until the first fault.

X~Exp(0.8)

Find the expected length until the first fault.

(a)