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MATH161 January 2008 Exam Solutions
All questions similar to seen exercises.
- (i) Median = obs. ( 20+1 2 ) = obs. 10.5 = (77 + 77)/2 = 77 [1 mark] LQ = obs. ( 20+1 4 ) = obs. 5.25 = 75 × 0 .75 + 75 × 0 .75 = 75 [1 mark] UQ = obs. (3/4) × (20 + 1) = obs. 15.75 = 83 × 0 .25 + 83 × 0 .75 = 83 [1 mark] (ii) Boxplot:
[3 marks] Boxplot suggests right skew, in that both box and whisker extend further to right than left. One outlier at 99, if this is excluded then right skewness is no longer so apparent in the whisker but still clear in the box. [2 marks]
- (i) For A, B independent, P (A ∪ B) = P (A) + P (B) − P (A)P (B). [2 marks]
(ii) For C, D, E mutually exclusive and exhaustive, P (C) = 1 − P (D) − P (E). [2 marks] (iii) Definition: P (G|F ) = P^ (G^ ∩^ F^ ) P (F ) [2 marks]
P (G|F ) =
P (G ∩ F )
P (F ) =^
P (G)P (F |G)
P (F ) [2 marks]
- Network reliability^ =^ P^ (A)^ ×^ P^ (B)^ ×^ P^ (C^ ∪^ D)^ ×^ P^ (E)
= 0. 9 × 0. 9 × (0.85 + 0. 85 − 0. 85 × 0 .85) × 0 .9 = 0. 7125975
[8 marks]
- (i) Var[X] = 4. 5 [1 mark]
(ii) P (X ≤ 2) =
(
- 50 0! +
) e−^4.^5 = (1 + 4.5 + 10.125) e−^4.^5 = 15. 625 × 0 .01111 = 0. 1736 [2 marks] (iii) P (X ≥ 1) = 1 − P (X = 0) = 1 − e−^4.^5 = 0. 9889 [2 marks] (iv) P (X = 0|X ≤ 2) = P^ (X^ = 0) P (X ≤ 2) = e
− 4. 5 15 .625e−^4.^5 = 0. 064 [3 marks]
- (i) With Z ∼ N (0, 1) then
P (X ≤ 26) = P
( Z ≤ 26 − 3 20
) = P (Z ≤ 2) = 0. 9772 [2 marks]
(ii) P (15 < X < 21) = P
3 < Z <^
) = P (− 1. 67 < Z < 0 .33) = P (Z < 0 .33) − (1 − P (Z < 1 .67)) = 0. 6293 − 1 + 0.9525 = 0. 5818 [3 marks] (iii) P (X < a) = 0. 6554 ⇒ P
( Z < a^ −^20 3
) = 0. 6554 ⇒ a^ −^20 3
⇒ a = 0. 4 × 3 + 20 = 21. 2 [3 marks]
- (i) 1 = ∫ (^2) 0 K^
( 2 w^3 − w^4 ) dw = K [ (w^4 /2) − (w^5 /5) ] 2 0 =^ K^ (8^ −^ (32/5)) = 8K/5, so that K = 5/8. [1 mark] Sketch of pdf:
[3 marks] (ii) (a) P (W < 0 .5) =
∫ (^0). 5 0 (5/8)^
( 2 w^3 − w^4
) dw = (5/8)
[ (w^4 /2) − (w^5 /5)
] 0. 5 0 = (5/8) ((1/32) − (1/160)) = (5/8)(1/40) = 1/64 = 0.015625. [2 marks] (b) P (W > 1) = ∫ (^2) 1 (5/8)^
( 2 w^3 − w^4 ) dw = (5/8) [ (w^4 /2) − (w^5 /5) ] 2 1 = (5/8) ((1/2) − (1/5)) = (5/8)(3/10) = 3/16 = 0.1875. [2 marks] (iii) E[W ] =
∫ (^2) 0 (5/8)^
( 2 w^4 − w^5
) dw = (5/8)
[ (2w^5 /5) − (w^6 /6)
] 2 0 = (5/8) ((64/5) − (32/3)) = (5/8) × (32/15) = 4/3. [2 marks] To find mode, f ′(w) = (5/8) ( 6 w^2 − 4 w^3 ) = (5/4)w^2 (3 − 2 w), stationary points at w = 0 and w = 3/2. From sketch of pdf in (i) above, mode is at w = 3/2. [2 marks] (iv) E
[ W 2
]
∫ (^2) 0 (5/8)^
( 2 w^5 − w^6
) dw = (5/8)
[ (w^6 /3) − (w^7 /7)
] 2 0 = (5/8) ((64/3) − (128/7)) = (5/8) × (64/21) = 40/21. [3 marks] Var[W ] = (40/21) − (4/3)^2 = (40/21) − (16/9) = 8/ 63 ≈ 0 .12698. [1 mark] (v) Distribution of W is skewed to the left, so would expect the median to lie to the left of the mode but not as far left as the mean, that is, expect 4/ 3 < Median < 3 /2. [4 marks]
- (i) Proportion in favour = (0 × 17 + 1 × 46 + 2 × 30 + 3 × 6 + 4 × 1)/400 = 128/400 = 0.32. [2 marks]
(ii) Each household consists of 4 voters, each of whom can be in one of two states (for or against), so that we have a fixed number of trials each with two possible outcomes, as required for a binomial distribution. The other condition required is that the 4 trials be independent of one another, which seems rather a dubious assumption – members of a single household are quite likely to hold similar political opinions. [4 marks] (iii) For X ∼ Bin(4, 0 .32), have
P (X = 0) =
( 4 0
)
- 684 = 0. 2138 ,
P (X = 1) =
( 4 1
)
- 683 = 0. 4025 ,
P (X = 2) =
( 4 2
)
- 682 = 0. 2841 ,
P (X = 3) =
( 4 3
)
- 681 = 0. 0891 ,
P (X = 4) =
( 4 4
)
- 680 = 0. 0105.
[4 marks] (iv) Expected values computed as Ei = 100P (X = i) are E 0 = 21. 4 , E 1 = 40. 3 , E 2 = 28. 3 , E 3 = 8. 9 , E 4 = 1.1. Then
X^2 = (17^ −^21 .4)
2
- 4
+ (46^ −^40 .3)
2
- 3
+ (30^ −^28 .3)
2
- 3
+ (6^ −^8 .9)
2
- 9
+ (1^ −^1 .1)
2
- 1 = 0 .9047 + 0.8062 + 0.1021 + 0.9449 + 0. 0091 = 2. 767 (More precise version: = 0 .8978 + 0.8223 + 0.0890 + 0.9520 + 0. 0023 = 2. 763 .)
[3 marks] To be compared with chi-squared distribution on 5 − 1 − 1 = 3 degrees of freedom. From tables, χ^23 (5%) = 7.815, so at the 5% level cannot reject the hypothesis that the data are from a binomial distribution. Binomial distribution appears to be appropriate for the given data. [3 marks] (v) To test whether binomial with p = 0.5 fits the data, would need to re-calculate values of P (X = i) using binomial probabiilities with n = 4, p = 0.5, then compute Ei values as before by multiplying probabilities by 100, compute X^2 test statistic using same formula, same observed values, but new Ei values, then compare X^2 value with chi- squared distribution now on 5 − 1 = 4 degrees of freedom, since no longer estimating any parameter values from the data. [4 marks]
- (i) Binomial (60, 0 .75). [3 marks]
(ii) Using normal approximation, have approximately X ∼ N (45, 11 .25). [2 marks] Hence with Z ∼ N (0, 1) we have
P (45 ≤ X ≤ 50) = P (44. 5 ≤ X ≤ 50 .5) ≈ P
≤ Z ≤ 50 √.^5 −^45
)
= P (− 0. 15 ≤ Z ≤ 1 .64)
= P (Z ≤ 1 .64) − (1 − P (Z ≤ 0 .15))
[4 marks] With n = 60, p = 0.75, have np = 45 > 5 and n(1 − p) = 15 > 5, so that normal approximation should be good. On the other hand, although n > 50, we don’t have p < 0 .1, so would not expect Poisson approximation to be good. [3 marks] (iii) Denoting by Y the number of regular internet users in the sample, have Y ∼ Bin(60, 0 .05), so approximate by Poisson(3). [1 mark] Have
P (Y < 5) ≈ e−^3
( 30 0! +
)
= e−^3 (1 + 3 + 4.5 + 4.5 + 3.375) = 0. 049787 × 16. 375 = 0. 8153
[4 marks] In this case, n = 60, p = 0.05, so np = 3 < 5 and hence would not expect normal approximation to be good. On the other hand, have n > 50 and p < 0 .1, so do expect Poisson approximation to be good. [3 marks]
