| 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(W|B) = 0.70 |
| a) P(W 3 B) = P(W|B) P(B) = 0.7 (0.6) = 0.42 | b) P(B|W) = 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 2-tail 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: l1 = l2 | Ha: l1 > l2 |
| 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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