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
Showing pages  1  -  2  of  7
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]
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
The preview of this document ends here! Please or to read the full document or to download it.
Document information
Embed this document:
Docsity is not optimized for the browser you're using. In order to have a better experience please switch to Google Chrome, Firefox, Internet Explorer 9+ or Safari! Download Google Chrome