2017 Stats Midterm Solutions
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### Question 1
#1.1 33
#1.2 11
#1.3 123.5
#1.4 41
#1.5
> 33 + 11*c(-2,2)
[1] 11 55
### Question 2
#2.1 .6
#2.2 2.6
> x1=14;y1=20
> x2=26;y2=55
> (y2-y1)/(x2-x1)
[1] 2.916667
#2.3 -18
> 35-2.6*20
[1] -17
#2.4 2.5
#2.5 16.9
> .6*2.5*11
[1] 16.5
#2.6
-18 + 2.6*18
[1] 28.8
### Question 3
#3.1 .8
#3.2
> 1*.2+2*.5+3*.3
[1] 2.1
#3.3
.2*(1-2.1)^2 + .5*(2-2.1)^2 + .3*(3-2.1)^2
[1] 0.49
> pv = c(.2,.5,.3)
> tv = 1:3
> sum(pv*tv)
[1] 2.1
> sum(pv*(tv-2.1)^2)
[1] 0.49
#3.4
> sqrt(.49)
[1] 0.7
#3.5
> 10+5*2.1
[1] 20.5
#3.6
> 5^2 * .49
[1] 12.25
#3.7
> sqrt(12.25)
[1] 3.5
> 5*.7
[1] 3.5
### Question 4
#4.1 .2
#4.2 .68
#4.3
> .56/(.56+.2)
[1] 0.7368421
#4.4 No. P(Y=1) != P(Y=1 | X=1).
#4.5 X ~ Bernoulli(.76)
#4.6 .76
#4.7
> .76*(1-.76)
[1] 0.1824
#4.8 No, they are not independent.
#4.9 given X=0, Y~Bernoulli(.5)
#4.10 Var(Y|X=0)=.25.
#4.11 Given X=0, var of a Bernoulli is maxed at p=.5.
#4.12 When you learn X=1 p is .74 but when you learn X=0, p=.5
so you can lean towards Y=1 when X=1 but when X=0 is is literally
a coin toss.
### Question 5
#5.1 .5
#5.2 .95
#5.3 .68
#5.4 .975
#5.5 .63
#5.6
> 1-.63
[1] 0.37
### Question 6
#6.1
There is no obvious pattern over time so IID Bern is plausible.
#6.2
> .3^3
[1] 0.027
#6.3
> .7^3
[1] 0.343
#6.4
same as #6.2 .027
#6.5
The D sequences such that T=1
are
1,0,0
0,1,0
0,0,1
each has prob .3*.7^2 = 0.147
So the prob is 3*.147 = 0.441
Note, T3 is the number of 1's out of 3 ``trials''.
Tn is the number of 1's out of n trials.
If we let p be the probability of a 1,
then the number of 1's out of n trials is
called a binomial random variable.
> dbinom(3,3,.3) #chance of the 3 1's out of 3 trials, p=.3
[1] 0.027
> dbinom(1,3,.3) #change of 1 1 out of 3 trials, p=.3
[1] 0.441
e.g suppose we toss a coin 10 times,
what is the distribution of the number of heads:
> x = 0:10
> px = dbinom(x,10,.5)
> cbind(x,px)
x px
[1,] 0 0.0009765625
[2,] 1 0.0097656250
[3,] 2 0.0439453125
[4,] 3 0.1171875000
[5,] 4 0.2050781250
[6,] 5 0.2460937500
[7,] 6 0.2050781250
[8,] 7 0.1171875000
[9,] 8 0.0439453125
[10,] 9 0.0097656250
[11,] 10 0.0009765625
e.g.
> .5^10
[1] 0.0009765625
### Question 7
#7.1
Assume I decide based on the expected value of my monetary
payout, if I promote I get 98 with prob .03 and -2 with prob .97.
> .03*98 + .97*(-2) = 1
So I promote.
#7.2
> .03*.53
[1] 0.0159
#7.3
> .97*.87
[1] 0.8439
#7.4
> .0159+.8439
[1] 0.8598
#7.5
> .0159/.8598
[1] 0.01849267
lower than the marginal prob of .03 which makes sense.
#7.6
> 100*.0185-2
[1] -0.15
> .0185*98 + (1-.0185)*(-2)
[1] -0.15
#7.7
> .03*.33
[1] 0.0099
> .97*.08
[1] 0.0776
> 0.0099/(0.0099+0.0776)
[1] 0.1131429
or just
.03*.33/(.03*.33 + .97*.08)
[1] 0.1131429
E(A) =
0.1131429*98 + (1-0.1131429)*(-2)
[1] 9.31429
should promote.