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]

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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

6. (a) (i) Ordered data:

Midazolam 200 272 346 359 366 371 426 497

Placebo 149 150 159 159 170 292 297 335

Midazolam: Median = (359 + 366)/2 = 362.5, LQ = 0.75 ×272 + 0.25 ×346 =

290.5, UQ = 0.25 ×371 + 0.75 ×426 = 412.25

Placebo: Median = (159+170)/2 = 164.5, LQ = 0.75×150+0.25×159 = 152.25,

UQ = 0.25 ×292 + 0.75 ×297 = 295.75 [4 marks]

Boxplots:

0 100 200 300 400 500

Activity level

Placebo

Midazolam

[4 marks]

(ii) The two distributions have similar spread, as indicated by inter-quartile range.

Midazolam distribution fairly symmetrical, Placebo distribution very right skewed.

Location much higher for Midazolam group, indicated by median and also by

whole box position. The eﬀect of midazolam appears to be to increase the activ-

ity levels of the rats. [4 marks]

(iii) Normal distribution would not be appropriate for activity levels of placebo group,

because distribution is very skewed. [2 marks]

(b) (i) No, because no upper limit on number of apples. [1 mark]

(ii) No, because height is continuous random variable. [1 mark]

(iii) No, because variable is ordinal. [1 mark]

(iv) Yes, continuous random variable, could plausibly be normally distributed. [1 mark]

(v) No, because no ﬁxed number of trials. [1 mark]

(vi) No, because time interval is continuous random variable. [1 mark]

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7. (i) Denote events R= Flier contains request, D= Flier disposed of in litter bin. Then

P(D) = P(D|R)P(R) + P(D|R)P(R)=0.09 ×0.49 + 0.05 ×0.51 = 0.0696 [2 marks]

(ii) P(R|D) = P(D|R)P(R)

P(D)=0.09 ×0.49

0.0696 = 147/232 = 0.6336 [4 marks]

(iii) X∼Binomial(10,0.0696). [3 marks]

Var[X] = 10 ×0.0696 ×0.9304 = 0.64756 [2 marks]

P(X≤3) = 10

00.06960×0.930410 +10

10.06961×0.93049+10

20.06962×0.93048+

10

30.06963×0.93047= 0.4861 + 0.3636 + 0.1224 + 0.0244 = 0.9965 [3 marks]

(iv) C= 0.1×10 + 0.15 ×(10 −X)+0.05 ×X= 2.5−0.1X[3 marks]

E[C]=2.5−0.1E[X]=2.5−0.1×10 ×0.0696 = 2.4304 [3 marks]

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