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MATH161 January 2009 Exam Solutions

All questions similar to seen exercises.

1. (i) Actresses Actors 8 6 5 2 9

9 5 5 4 3 3 2 6 8 9 5 4 0 3 5 6 7

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) A and B not 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) A and C are not independent, because P (C∩(A∪B)) = 0, so in particular P (C∩A) = 0 6= P (C)× P (A). [1 mark]

3. (i) ∫ 3 0 K

( 39− 4x− 3x2

) dx = K

[ 39x− 2x2 − x3

]3 0 = 72K, so K = 1/72 [2 marks]

(ii) E[X] = ∫ 3 0 K

( 39x− 4x2 − 3x3

) dx = K

[ (39/2)x2 − (4/3)x3 − (3/4)x4

]3 0

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

(iii) F (x) = ∫ x 0 K

( 39− 4t− 3t2

) dt = K

[ 39t− 2t2 − t3

]x 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 S denote number of sales in 12 calls.

(i) E[S] = 12× 0.08 = 0.96 [2 marks]

(ii) P (S ≤ 2) = (12

0

) 0.080 × 0.9212 +

(12 1

) 0.081 × 0.9211 +

(12 2

) 0.082 × 0.9210

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

(iii) P (S ≥ 2) = 1− P (S ≤ 1) = 1− (12

0

) 0.080 × 0.9212 −

(12 1

) 0.081 × 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: χ22(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]

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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 0

) 0.06960 × 0.930410 +

(10 1

) 0.06961 × 0.93049 +

(10 2

) 0.06962 × 0.93048 +(10

3

) 0.06963 × 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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8. (i) Plot:

0 1 2 3 4 5 6 7 8 0

4

8

12

16

20

24

Number of colonies

Frequency

[4 marks]

(ii) Mean = (0×12+1×10+2×24+3×12+4×10+5×6+6×12+7×12+8×2)/100 = 336/100 = 3.36 [2 marks]

(iv) Counts might be reasonably expected to follow a Poisson distribution, being a discrete random variable, not bounded above, counting number of events in some fixed region (the agar plate). [2 marks]

(iii) Expected values calculated as Ei = 100× (3.36i/i!)× e−3.36 [1 mark]

Number of colonies 0 1 2 3 4 5 6 7 ≥ 8 Expected frequency 3.47 11.67 19.61 21.96 18.45 12.40 6.94 3.33 2.17

[3 marks] Since 3 of the 9 expected values are less than 5, which is more than 20% of categories, need to combine the last two categories, giving

Number of colonies 0 1 2 3 4 5 6 ≥ 7 Observed frequency 12 10 24 12 10 6 12 14 Expected frequency 3.47 11.67 19.61 21.96 18.45 12.40 6.94 5.50

[1 mark]

Now X2 = 7∑

i=0

(Oi − Ei)2

Ei =

(12− 3.47)2

3.47 + · · ·+ (14− 5.50)

2

5.50 = 20.93 + 0.24 + 0.98 + 4.52 + 3.87 + 3.30 + 3.69 + 13.12 = 50.64 [3 marks]

Have 8 categories, 1 parameter estimated from data, so degrees of freedom = 8− 1− 1 = 6. [1 mark]

Critical value χ26(5%) = 12.59 [1 mark]

Therefore evidence at the 5% level that the Poisson distribution is not appropriate for the given data. [2 marks]

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9. (i) Denoting X = Number of calls in half-hour period, then X ∼ Poisson(3.6), so

P (X > 3) = 1− P (X ≤ 3) = 1− e−3.6 (

3.60

0! +

3.61

1! +

3.62

2! +

3.63

3!

) = 1−0.02732× (1 + 3.6 + 6.48 + 7.776) = 1−0.02732×18.856 = 1−0.5152 = 0.4848 [4 marks]

(ii) P (X = 5|X > 3) = P (X = 5 ∩X > 3) P (X > 3)

= P (X = 5) P (X > 3)

= e−3.6 × 3.65/5!

0.4848 =

0.1377 0.4848

= 0.2840 [4 marks]

(iii) Denote U = Number of calls between 10:00 and 10:20, and V = Number of calls between 10:20 and 10:30.

Then U, V are independent with U ∼ Poisson(2.4), V ∼ Poisson(1.2), and U + V ∼ Poisson(3.6).

Hence P (V = 1|U + V = 5) = P (V = 1 ∩ U + V = 5) P (U + V = 5)

= P (V = 1 ∩ U = 4) P (U + V = 5)

= P (V = 1)× P (U = 4)

P (U + V = 5) =

(e−1.2 × 1.21/1!)× (e−2.4 × 2.44/4!) e−3.6 × 3.65/5!

= 0.3614× 0.1254

0.1377 = 0.3292 [5 marks]

(iv) Denoting by Y the number of calls in a 4 hour period, then Y ∼ Poisson(1.2 × 6 × 4) = Poisson(28.8). For λ > 15, can approximate Poisson(λ) by Normal(λ, λ), so approximate Y by N(28.8, 28.8). [3 marks]

Denoting by Z a standard normal variate, then P (Y > 25) = P (Y ≥ 25.5) ≈ P

( Z ≥ 25.5− 28.8√

28.8

) = P (Z ≥ −0.6149) = P (Z ≤ 0.61) = 0.7291 [4 marks]

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10. (a) Denote by H the height of a randomly chosen man (in cm), so H ∼ N ( 175, 7.112

) ,

and by Z a standard normal variate.

(i) P (H > 192) = P ( Z >

192− 175 7.11

) = P (Z > 2.39) = 1− P (Z ≤ 2.39)

= 1 − 0.99158 = 0.00842, so that 0.842% of men are too tall to fit through the doorway. [3 marks]

(ii) Denoting by M the number of men in a sample of 10 who are too tall to fit through the doorway, then M ∼ Binomial(10, 0.00842). Hence P (M = 2) =

(10 2

) 0.008422 × 0.991588 = 0.00298 [3 marks]

(iii) Require h such that

P (H > h) = 0.02

P

( Z >

h− 175 7.11

) = 0.02

P

( Z ≤ h− 175

7.11

) = 0.98

h− 175 7.11

= 2.05

h = 2.05× 7.11 + 175 = 189.6

Required height is 189.6 cm. [3 marks]

(b) (i) 95% CI is x̄± 1.96σ/ √ n = 162± 1.96× 6.35/

√ 100 = 162± 1.2446

= [160.76, 163.24] cm. [6 marks]

(ii) The value 175 lies outside the interval computed in part (i), so at the 5% level we can reject H0 : µ = 175 in favour of H1 : µ 6= 175. There is evidence that the mean height of women differs from 175 cm. [3 marks]

(iii) With a 10% significance level, we would not require such strong evidence to reject 175 cm, and the corresponding 90% confidence interval would be narrower than the 95% confidence interval of (i) above. That is, if 175 cm is rejected at the 5% level, it would certainly be rejected at the 10% level. The conclusion of part (ii) would not change. [2 marks]

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