Let us suppose that some article investigated the probability of corrosion of steel reinforcement in concrete...

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Let us suppose that some article investigated the probability ofcorrosion of steel reinforcement in concrete structures. It isestimated that the probability of corrosion is 0.19 under specificvalues of half-cell potential and concrete resistivity. The risk ofcorrosion in five independent grids of a building with these valuesof half-cell potential and concrete resistivity. Let the randomvariable X denote number of grids with corrosion in this building.Determine the cumulative distribution function for the randomvariable 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 & Explanation Solved by verified expert
4.4 Ratings (883 Votes)

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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