Answer:
P(Y=1|X=3)=0.125
Step-by-step explanation:
Given :
p(1,1)=0
p(2,1)=0.1
p(3,1)=0.05
p(1,2)=0.05
p(2,2)=0.3
p(3,2)=0.1
p(1,3)=0.05
p(2,3)=0.1
p(3,3)=0.25
Now we are supposed to find the conditional mass function of Y given X=3 : P(Y=1|X=3)
P(X=3) = P(X=3,Y=1)+P(X=3,Y=2) +P(X=3,Y=3)
P(X=3)=p(3,1) +p(3,2) +p(3,3)
P(X=3)=0.05+0.1+0.25=0.4

Hence P(Y=1|X=3)=0.125
1b,
2b,
3a, which grade is this?
The empirical rule states that approximately 68/95/99.7% of a normal distribution lies within 1/2/3 standard deviations. So the answer is 68%.
It is an over estimate because 5.82-4.29=1.53 it is closer to 1.5 not 2
Answer: I think it's 24
Step-by-step explanation: