a) We have that
pX(x)≥0,x=1,2,3,4,5i=1∑5pX(xi)=1Given that pX(x)=cx,x=1,2,3,4,5. Then
c⋅1+c⋅2+c⋅3+c⋅4+c⋅5=115⋅c=1c=151
b) The probability that at most three forms are required
P(X≤3)=P(X=1)+P(X=2)+P(X=3)==151⋅1+151⋅2+151⋅3=52
c) The probability that between two and four forms (inclusive) are required
P(2≤X≤4)=P(X=2)+P(X=3)+P(X=4)==151⋅2+151⋅3+151⋅4=53 d) Check that i=1∑5pX(xi)=1
5012+5022+5032+5042+5052=1011=1 Therefore, no. A probability distribution of x
pX(x)=50x2,x=1,2,3,4,5 could not be.
No.
Comments
Leave a comment