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) = = =

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2) P( loss ) = 1 - 5/9 = 4/9 so odds against a loss are 5:4

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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.

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4)a) S P(x) = 1 and 0 [ P(x) [ 1, ¼ x .

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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

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 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, P( E < | 5 | ) = at least 75%

 b) the Central Limit Theorem sets z = ! 2, We want 2 [P( z < 2)] = 2 (0.4772) = 95.44%

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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

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7) a)

 The mean l = 48/6 = 8 the standard deviation r = 3.42

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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

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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

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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

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8)

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9) a) x = 375, n = 500 so p = 375/500 = 0.75
r = [np(1 - p)] ½ = 9.68

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b) .

So 0.718 [ p [ 0.782

The 2-tail confidence interval is 0.718 [ p [ 0.782

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10) s = \$3,000, E = \$200 n = ?

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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.

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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.

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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.

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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.

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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.

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TOTAL (100)

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