b0=
-b1
=532/10-2.016·311/10=-9.5so
y=-9.5+2.016x
| x | 23 | 45 | 17 | 31 | 39 | 29 | 27 | 35 | 35 | 30 |
| y | 40 | 100 | 35 | 55 | 45 | 25 | 40 | 60 | 72 | 60 |
n=10
∑x=311
∑x2=10245
∑y=532
∑y2=32484
∑xy=17700
so Sxx=∑x2-(∑x)2/n=10245-3112/10=572.9
Sxy=∑xy-∑x·∑y/n=17700-532·311/10=1154.8
b1=Sxy/Sxx=1154.8/572.9=2.016
b0=-b1=532/10-2.016·311/10=-9.5
so
y=-9.5+2.016x
Problem 2 A box contains 4 blue balls and 10 red balls.
a) One ball is picked from the box. The rv X is 1 if the ball is blue, 0 if it is red. Find the probability mass function of X.
| x | 0 | 1 |
|---|---|---|
| P(X=x) | 5/7 | 2/7 |
b) Say the first ball is put aside and a second ball ball is picked from the box. Let the rv X be the number of balls picked that are blue. Find the probability mass function of X.
P(X=0)=P(both red)=10/14·9/13=0.4945
P(X=2)=P(both blue)=4/14·3/13=0.0659
P(X=1)=1-P(X=0or2)=1-(0.4945+0.0659)=0.4396
| x | 0 | 1 | 2 |
|---|---|---|---|
| P(X=x) | 0.4945 | 0.4396 | 0.0659 |
Problem 3 It is known that in a certain population 45% of the people are male.
a) If we randomly select 75 people from this population, what is the mean and the standard deviation of the number of females in the sample?
X~Bin(75,0.55), so m=np=75·0.55=41.25 and s=√(npq)=√(74·0.550.45) = 4.31
b) if we randomly select one person after the other, what is the probability that the first male selected is the fourth person?
X~G(0.45), so P(X=4)=0.45·0.553 = 0.0749
c) if we randomly select 8 people from this population, what is the probability that exactly 5 of them are female?
X~Bin(8,0.55), so P(X=5)=P(X≤5)-P(X≤4)=0.780-0.523=0.257 (from the tables)
Problem 4
a) S={(r,r),(r,b),(r,g),(b,b),(b,g),(g,g)}
b) P=3/6=1/2
c) same as above. Now outcomes are not equally likely.
Problem 5
a) Dataset 1: r=-0.75
Dataset 2: r=0
Dataset 3: r=0.99
Dataset 4: r=0.85
b) Dataset 1: we have points (0,50) and (100,28), so b1=(28-50)/(100-0)=-0.32 and b0=50-(-0.32)·0=50, so y=50-0.32x