b) We have the formula
so if we knew Sxx and Syy we could find r. But
the standard deviation of x is
s=√[Sxx/(n-1)]
so
Sxx=(n-1)·s2
Also roughly s=range/4=100/4=25, so
Sxx=(21-1)·252=12500
and for y
s=(210-40)/4=42.5
Syy=(21-1)·42.52=36125
and now
r=(-1.65)·√(12500/36125)=-0.97
(The true value is -0.98)
Problem 2
a) not possible, the point (
,
) is always on the line, but 23.5+17.9*5.7 = 125.53, not 110.9
b) not possible, r is always between -1 and 1
c) not possible, b1 and r always have the same sign, but
y=-1.5+0.64x implies that b1>0, so r=-0.55 is impossible
d) not possible, y=5.6+9.1x says that cor(x,y)>0, x=7.8-4.4y says that cor(y,x)<0, but cor(x,y)=cor(y,x)
Problem 3
a) S={(r,r),(r,b), (r,g), (b,b), (b,g)}
b) no, because the is only one green ball for example (r,b) is more likely than (r,g)
Problem 4
a) If you replace "1st die" with "1st roll" and "2nd die" with "2nd roll" you see that the sample space is the same as the one discussed in class here and we see there are 30 pairs were the first number is different from the second. So P("Rolls are different") = 30/36 = 5/6
b) Looking at the sample space we see that there are 6 pairs were the second number is a 6. So P("Second rolls is a six") = 6/36 = 1/6
c) A = "The two rolls differ by more than 2" = {(1,4), (1,5), (1,6), (2,5), (2,6), (3,6), (4,1), (5,1), (5,2), (6,1), (6,2), (6,3)}, so n(A)=12, so P(A) = 12/36 = 1/3.
d) A="second number is even", so n(A)=18, and P(A)=n(A)/n(S)=18/36 = 1/2
e) A="product is at most 5" = {(1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (3,1), (4,1), (5,1)}, so n(A)=10, so P(A)=10/36=5/18