Lösungen / Statistik 1/03

Remove["Global`*"]

1.

<< Statistics`DescriptiveStatistics`

a = {1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 
          3, 4, 4, 4, 5, 5, 6, 6, 7, 8, 9, 10};
b = {1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 
          3, 4, 4, 4, 5, 5, 6, 6, 7, 8, 9, 10, 500001};

LocationReport[a] // N

LocationReport[b] // N

Mean[b] - Mean[a] // N

Median[b] - Median[a] // N

2.

\(\(c = {{2, 3}, {3, 5}, {1, 3}, {4, 6}, {1, 6}, {1, 6}, {2, 2}, {4, 
            6}, {4, 6}, {5, 6}, {1, 1}, {2, 6}, 
{2, 5}, {1, 1}, {2, 5}, {1, 
            4}};\)\)

<< Statistics`DataManipulation`

Frequencies[c]

hG = Length[c]

hA = 10;  hB = 5;  hAandB = 4;

hAorB = hA + hB - hAandB

2.

htotal = 6^6

a

hA = 2 ((1*1*6)*3*6^3)

hA/htotal

hA/htotal // N

b

hB = 2 ((1*1*5)*3*6^3)

hB/htotal

hB/htotal // N

c

hC = htotal - hA + hB

hC/htotal // N

4.

g = 100; a = 50; b = 45; aUnionb = 80; 
    aIntersb = 50 + 45 - 80

5.

r1 = 7/10; r2 = 6/9;

r12 = r1 r2

N[%]

6.

p = 4/36 3/35

N[%]

7.

?Bin*

Total Möglichkeiten / Total possibilités

total = Binomial[42, 6]

Exakt richtige 4 der 6 heisst zudem 2 falsche von 38 / exactement 4 de 6 juste signifie en plus 2 fauts de 38

e4 = Binomial[38, 2]

r4 = Binomial[6, 4]

e4 r4/total

N[%]

Exakt richtige 5 der 6 heisst zudem 1 falsche von 37/ exactement 5 de 6 juste  signifie en plus 1 fauts d 37

e5 = Binomial[37, 1]

r5 = Binomial[6, 5]

e5 r5/total

N[%]

Exakt richtige 6 der 6 heisst 0 falsche von 36/ exactement 6 de 6 juste dsignifie 1 fauts d 36

e6 = Binomial[36, 1]

r6 = Binomial[6, 6]

e6 r6/total

N[%]

Mindestens 4 richtige / au moins 4 justes

(e4 r4 +  e6 r6 + e6 r6)/total

N[%]