Question

3. A metal fabricating plant currently has five major pieces under contract each with a deadline for completion. Let X be the number of pieces completed by their deadlines, and suppose it's PMF p(x) is given by x 0 1 2 3 4 5 p(x) .05 .1 .15 .25 .35 .1 (a) Find and plot the CDF of X. (b) Use the CDF to find the probability that between one and four pieces, inclusive, are completed by their deadline

Answer 1

a) The cumulative distribution function would be given by:

x 0 1 2 3 4 5

F(X) 0.05 0.15 0.30 0.55 0.9 1

b)
`P(1 \leq X \leq 4) = F(4) -F(0) =0.9-0.05 = 0.85`

And replacing we got:

`P(1 \leq X \leq 4) =0.85`

Explanation:

For this case we have the following probability distribution function given:

x 0 1 2 3 4 5

P(X) 0.05 0.1 0.15 0.25 0.35 0.1

We satisfy the conditions in order to have a probability distribution:

1)
`\sum_(i=1)^n P(X_i)=1`

2)
`P(X_i) \geq 0, i=1,2,..,n`

Part a

The cumulative distribution function would be given by:

x 0 1 2 3 4 5

F(X) 0.05 0.15 0.30 0.55 0.9 1

Part b

For this case we want to find this probability:

`P(1 \leq X \leq 4) = F(4) -F(0) =0.9-0.05 = 0.85`

And replacing we got:

`P(1 \leq X \leq 4) =0.85`