Solution Regular Exam
Marks
Q1
1 (a)
Mean =
11.9 $bn
2
SD =
1.3 $bn
1 (b)
Mean =
11.9 $bn
Covar =
0.28
3
SD =
1.5 $bn
(c )
Tchebyshev
P(within 2 SDs of mean) >
0.75
P(outside 2 SDs of mean) <
0.25
P(< mean + 2 * SD) >
0.88
3
Mean + 2 * SD =
10 $bn
Q2
(a)
Binomial
p=
0.46
n=
10
3
P(x=6) =
0.17
(b)
Normal Approx to Binomial
n=
100
np =
46
n(1p) =
54
Since these are >5, we can use the Normal Approximation
Mean =
46
SD =
4.98
z=
2.71
Prob from table =
0.5
5
P(>60) =
0
1 (c )
Maybe
There is a 17% chance of getting 6/10 satisfied with ineffective training
1
This % is not small enough to prove the training was effective
Marks
Q3
(a)
1
P(X>0)=P(z>15/25)=P(z>0.6)=
1
0.5+0.2257=0.7257
(b)
1
P(x>y)=P(xy>0)
1
E=5
1
s.d. = sqrt(25^2+35^2)
43.01
1
P(XY>0)=P(z>5/43.01161)=P(z>0.1162)=
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1
0.50.0478=
0.45
(c )
Pr(X>R)=0.3
1
Pr(z>(R20)/35)=0.3
1
(R20)/35=0.525
1
R=
38.38
Q4
(a)
28,000
29,000
27,000
28,000
25,000
35,000
24,000
28,000
1
mean
28,100
10000
810000
1210000
10000
9610000 47610000 16810000
10000
1
sd
5342.7
1
need to find prob(x_bar>28100)
1
this is equivalent to P(t
4.79 )
1
with d.f. = 9
1
This probability is < 0.005, i.e., very small. So, the official is wrong
(b)
1
x_bar=N(20,000, 8,000/sqrt(60))
s.d. =
1032.8
1
P(x_bar<18,000)=P(z<(18,00020,000)/1033)=P(z<1.94)=
1
0.50.4738=0.0262
(c )
1
We can use CLT and the result will be approximately the same
Qu.#5
10
2
a.
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 Winter '09
 na
 Probability theory, consultant, ml, tles

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