Best Actress - Mathematics - Solved Exam, Exams for Mathematics. Banasthali Vidyapith

Mathematics

Description: This is the Past Solved Exam of Mathematics which includes Vector, Scalar Product, Vector Product, Speed, Acceleration, Particle, Position, Magnitude, Same Direction, Scalar Field etc. Key important points are: Best Actress, Particular, Best Actor, Data, Median Age, Calculate, Comment, Common Difference, Geometric Serie, Differentiate
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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(AB) = P(A)×P(B|A)=0.4×0.5 = 0.2 [1 mark]
(ii) P(B) = P(AB) + P(AB)=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(AB). [1 marks]
(iv) Since P(C(AB)) = 0, then P(ABC) = P(C) + P(AB)
=P(C) + P(A) + P(B)P(AB) = 0.2+0.4+0.55 0.2=0.95 [3 marks]
(v) Aand Care not independent, because P(C(AB)) = 0, so in particular P(CA) =
06=P(C)×P(A). [1 mark]
3. (i) R3
0K39 4x3x2dx =K39x2x2x33
0= 72K, so K= 1/72 [2 marks]
(ii) E[X] = R3
0K39x4x23x3dx =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 4t3t2dt =K39t2t2t3x
0=39x2x2x3/72 [2 marks]
(iv) F(1) = (39 21) /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(S2) = 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(S2) = 1 P(S1) = 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]
3
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) XBinomial(10,0.0696). [3 marks]
Var[X] = 10 ×0.0696 ×0.9304 = 0.64756 [2 marks]
P(X3) = 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= 2.50.1X[3 marks]
E[C]=2.50.1E[X]=2.50.1×10 ×0.0696 = 2.4304 [3 marks]
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