Question

Let us suppose that some article investigated the probability of corrosion of steel reinforcement in concrete structures. It is estimated that the probability of corrosion is 0.19 under specific values of half-cell potential and concrete resistivity. The risk of corrosion in five independent grids of a building with these values of half-cell potential and concrete resistivity. Let the random variable X denote number of grids with corrosion in this building. Determine the cumulative distribution function for the random variable X.

Round your answers to five decimal places (e.g. 98.76543).

f(x)=           with x < 0

f(x)=             with 0 <= x < 1

f(x)=             with 1 <= x < 2

f(x)=             with 2 <= x < 3

f(x)=             with 3 <= x < 4

f(x)=             with 4 <= x < 5

f(x)=             with 5 <= x

Answer 1

using Excel

x	p	cmf
0	0.348678	0.348678
1	0.408944	0.757622
2	0.19185	0.949472
3	0.045002	0.994474
4	0.005278	0.999752
5	0.000248	1

x	p	cmf
0	=BINOM.DIST(A2,5,0.19,0)	=B2
1	=BINOM.DIST(A3,5,0.19,0)	=C2+B3
2	=BINOM.DIST(A4,5,0.19,0)	=C3+B4
3	=BINOM.DIST(A5,5,0.19,0)	=C4+B5
4	=BINOM.DIST(A6,5,0.19,0)	=C5+B6
5	=BINOM.DIST(A7,5,0.19,0)	=C6+B7

f(x)=     0      with x < 0

f(x)=     0.34867844        with 0 <= x < 1

f(x)=         0.75762229    with 1 <= x < 2

f(x)=         0.949472491    with 2 <= x < 3

f(x)=     0.99447439        with 3 <= x < 4

f(x)=        0.99975239     with 4 <= x < 5

f(x)=           1 with 5 <= x

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