STATISTICS FINAL EXAM SOLUTIONS 
Write neatly, big enough to see, show the formulae you use:
1)
a) P( 3 Queens)  b) P( 2 ©, 1 ¨)  c) P( 6, 2 Kings) 
=  =  = 
(6)
2) P( loss ) = 1  5/9 = 4/9 so odds against a loss are 5:4
(4)
3)
P(W) = 0.50  P(B) = 0.60  P(WB) = 0.70 
a) P(W 3 B) = P(WB) P(B) = 0.7 (0.6) = 0.42  b) P(BW) = P(B 3 W)/P(W) = 0.42/0.5 = 0.84  
c) P(B 4 W) = P(B) + P(W)  P(B 3 W) = 0.68  d) P(B 4 W)^{ } = 1  P(B 4 W) = 1  0.68 = 0.32  
e) If (B 3 W)= P(B) P(W), then B and W are independent events But 0.42 ! 0.30, so B and W are not independent events. 
(10)
4)a) S P(x) = 1 and 0 [ P(x) [ 1, ¼ x .
(3)
b) both conditions are satisfied so this is a probability distribution.
f(0) = 1/16  f(1) = 4/16  f(2) = 6/16  f(3) = 4/16  f(4) = 1/16 
x  P(x) 
0  1/16 = .0625 
1  4/16 = .25 
2  6/16 = .375 
3  4/16 = .25 
4  1/16 = .0625 
Total  1 
(8)
5) n = 64,  r_{ x} = 20/8 = 2.5  k = E / r_{ x} = 2 
a) Chebyshev says P( E <  5  ) = at least 1  (1/k)^{ 2},

b) the Central Limit Theorem sets z = ! 2, We want 2 [P( z < 2)] = 2 (0.4772) = 95.44% 
(6)
6)
p = 0.25  n = 90  µ = np = 22.5  r = [np(1  p)]^{ ½} = 4.11 
a) for 24 cars, we use x = 23.5
so,
z_{ 0.24} = 0.0948, so P(x < 24) = 0.5 + 0.0948 = 0.5948
b) for 20 cars, we use x = 19.5
so,
z_{ 0.73} = 0.2673, so P(20 < x < 23) = 0.0948 + 0.2673 = 0.3621
(6)
7) a)
The mean l = 48/6 = 8  the standard deviation r = 3.42 
(2)
b) 15 random samples (n = 2) and l their means.
(3, 5) l = 4  (3, 13) l = 8  (5, 13) l = 9  (9, 11) l = 10 
(3, 7) l = 5  (5, 7) l = 6  (7, 9) l = 8  (9, 13) l = 11 
(3, 9) l = 6  (5, 9) l = 7  (7, 11) l = 9  (11, 13) l = 12 
(3, 11) l = 7  (5, 11) l = 8  (7, 13) l = 10 
(5)
c) the sampling distribution of the mean.
l  P(l) 
4  1/15 
5  1/15 
6  2/15 
7  2/15 
8  3/15 
9  2/15 
10  2/15 
11  1/15 
12  1/15 
Total  15/15 = 1 
(2)
d)
mean l = S [x $ P(x) ] = 8  variance = r^{ 2} = = 4.67 
so, the means are the same = 8  r^{ } = 2.16 < 3.42 
(4)
8)
(5)
9) a) x = 375, n = 500 so p = 375/500 = 0.75
r = [np(1  p)]^{ ½} = 9.68
(2)
b) .
So 0.718 [ p [ 0.782
The 2tail confidence interval is 0.718 [ p [ 0.782
(5)
10) s = $3,000, E = $200 n = ?
(4)
11)
Hypothesis Test on Mean  
Ho: l = 56  Ha: l < 56 
Level of significance: a = 5 %  Criteria: if t < 1.796, reject Ho 
Statistic:  
Conclusion: we must reject the null hypothesis. She can conclude that l < 56. 
(6)
12)
Hypothesis Test on Difference of Means  
Ho: l_{1} = l_{2}  Ha: l_{1} > l_{2} 
Level of significance: a = 1 %  Criteria: if z > 2.33, reject Ho 
Statistic:  
Conclusion: we must reject the null hypothesis Ho. There is enough evidence to support the claim. 
(6)
13) numbers in ( ) are expected values e.
VANILLA  CHOCOLATE  STRAWBERRY 
17 (20)  24 (20)  19 (20) 
Chi Square Hypothesis Test  
Ho: There is no flavour preference  Ha: There is a flavour preference 
Level of significance: a = 1%  Criteria: if x^{ 2} > 9.210 heave Ho. 
Statistic:  
Conclusion: we cannot reject Ho. It seems there is no flavour preference. 
(5)
14)
Chi Square Hypothesis Test  
Ho: preference is independent of program  Ha: preference is dependent on program 
Level of significance: a = 5 %  Criteria: if x^{ 2} > 5.991 heave Ho. 
Statistic:  
Conclusion: we cannot reject Ho. It seems preference is independent of program. 
(6)
15)
Chi Square Hypothesis Test  
Ho: proportions are the same  Ha: proportions are different 
Level of significance: a = 1%  Criteria: if x^{ 2} > 13.277 heave Ho. 
Statistic:  
Conclusion: we must reject Ho. It seems colour distribution is different from expected. 
(5)
TOTAL (100)
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