Answer:6 36 2.449
Step-by-step explanation: 36 square
Where are the functions?
your question is incomplete?
The outlier (61) is at the low end of the data set, but doesn't affect the mean by a lot, so ...
The mean is centered among the other numbers in both sets of data.
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The mean without the outlier is 114. With the outlier, it is 107.4. The lower quartile is 108, so the mean does get moved outside the "box" of the box-and-whisker plot of the data set without the outlier.
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
![P(Y=1|X=3)=\frac{P(X=3,Y=1)}{P(X=3)} =\frac{p(3,1)}{P(X=3)}=\frac{0.05}{0.4}= 0.125](https://tex.z-dn.net/?f=P%28Y%3D1%7CX%3D3%29%3D%5Cfrac%7BP%28X%3D3%2CY%3D1%29%7D%7BP%28X%3D3%29%7D%20%3D%5Cfrac%7Bp%283%2C1%29%7D%7BP%28X%3D3%29%7D%3D%5Cfrac%7B0.05%7D%7B0.4%7D%3D%200.125)
Hence P(Y=1|X=3)=0.125