x P(x) -3 0.15 -2 0.24 -1 p1 0 p2 1 0.12 2 0.24 a. p1 = 0.14, p2 = 0.39 b. p1 = –0.1, p2 = 0.35 c. p1 = 0.25, p2= 0.15 d. p1= 0.1, p2 = 0.15
x P(x)
-3 0.15
-2 0.24
-1 p1
0 p2
1 0.12
2 0.24
The sum of the P(x) must be 1
0.15+0.24+p1+p2+0.12+0.24 = 1
0.75+p1+p2 = 1
p1+p2 = 0.25
So pi+p2 must equal 0.25
a. p1 = 0.14, p2 = 0.39 <-- that's not it because 0.14+0.39 = 0.53, not 0.25
b. p1 = –0.1, p2 = 0.35 <-- that's not it because a negative number
cannot be a probability.
c. p1 = 0.25, p2= 0.15 <-- that's not it because 0.25+0.15 = 0.40, not 0.25
d. p1= 0.1, p2 = 0.15 <-- that's it because 0.1+0.15 = 0.25
Edwin
Determine the required value of the missing probability to make the distribution a...(answered by ikleyn,Edwin McCravy)