MATH161 January 2009 Exam Solutions

All questions similar to seen exercises.

1. (i) Actresses Actors

8 6 5 2 9

955433268

95403567

5

6 0

where 2|9 represents 29 years of age. [4 marks]

(ii) Sample size 10, so median is observation number 5.5.

Actress median = (34 + 35)/2 = 34.5 years

Actor median = (40 + 43)/2 = 41.5 years [2 marks]

(iii) Actresses seem overall younger, as indicated by lower median age. Actress age dis-

tribution also slightly skewed to left, whereas actor distribution skewed to right with

one outlier at age 60. [2 marks]

2. (i) P(A∩B) = P(A)×P(B|A)=0.4×0.5 = 0.2 [1 mark]

(ii) P(B) = P(A∩B) + P(A∩B)=0.2+0.35 = 0.55 [2 marks]

(iii) Aand Bnot independent, because P(A)×P(B)=0.4×0.55 = 0.22 6=P(A∩B). [1 marks]

(iv) Since P(C∩(A∪B)) = 0, then P(A∪B∪C) = P(C) + P(A∪B)

=P(C) + P(A) + P(B)−P(A∩B) = 0.2+0.4+0.55 −0.2=0.95 [3 marks]

(v) Aand Care not independent, because P(C∩(A∪B)) = 0, so in particular P(C∩A) =

06=P(C)×P(A). [1 mark]

3. (i) R3

0K39 −4x−3x2dx =K39x−2x2−x33

0= 72K, so K= 1/72 [2 marks]

(ii) E[X] = R3

0K39x−4x2−3x3dx =K(39/2)x2−(4/3)x3−(3/4)x43

0

=K((351/2) −36 −(243/4)) = (315/4)K= 315/288 = 1.09375 [3 marks]

(iii) F(x) = Rx

0K39 −4t−3t2dt =K39t−2t2−t3x

0=39x−2x2−x3/72 [2 marks]

(iv) F(1) = (39 −2−1) /72 = 36/72 = 1/2, so median value is at m= 1 [1 mark]

4. Let Sdenote number of sales in 12 calls.

(i) E[S] = 12 ×0.08 = 0.96 [2 marks]

(ii) P(S≤2) = 12

00.080×0.9212 +12

10.081×0.9211 +12

20.082×0.9210

= 0.3677 + 0.3837 + 0.1835 = 0.9348 [2 marks]

(iii) P(S≥2) = 1 −P(S≤1) = 1 −12

00.080×0.9212 −12

10.081×0.9211

= 1 −0.3677 −0.3837 = 0.2487 [2 marks]

(iv) P(Fifth call is ﬁrst sale) = 0.924×0.08 = 0.0573 [2 marks]

1

5. Expected values:

F S T Total

S 122.99 107.85 267.16 498

D 202.01 177.15 438.84 818

Total 325 285 706 1316

[3 marks]

Test statistic: X2=(202 −122.99)2

122.99 +(118 −107.85)2

107.85 +(178 −267.16)2

267.16

+(123 −202.01)2

202.01 +(167 −177.15)2

177.15 +(528 −438.84)2

438.84

= 50.76 + 0.96 + 29.76 + 30.91 + 0.58 + 18.12 = 131.08 [2 marks]

Critical value: χ2

2(5%) = 5.991

Test statistic much larger than critical value, so there is evidence at the 5% level of an

association between survival chance and class of ticket. [3 marks]

2

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