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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

Respuesta :

Answer:

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) [tex] P(1 \leq X \leq 4) = F(4) -F(0) =0.9-0.05 = 0.85 [/tex]

And replacing we got:

[tex]P(1 \leq X \leq 4) =0.85[/tex]

Step-by-step 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) [tex] \sum_{i=1}^n P(X_i)=1[/tex]

2) [tex] P(X_i) \geq 0, i=1,2,..,n[/tex]

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:

[tex] P(1 \leq X \leq 4) = F(4) -F(0) =0.9-0.05 = 0.85 [/tex]

And replacing we got:

[tex]P(1 \leq X \leq 4) =0.85[/tex]