Statistics Ii Worksheet #1 ( Solutions)

UNIVERSITY OF TECHNOLOGY, JAMAICA STATISTICS II (STA 2004) Worksheet #1 (Solutions) Question # 1 (i) a = 0.1 (ii) P ( 1 ≤ X ≤ 3) = P ( X = 1) + P ( X = 2 ) + P ( X = 3) = 0.2 + 0.25 + 0.4 = 0.85 (iii)

P ( X > 2 ) = P ( X = 3) + P ( X = 4 ) + P ( X = 5 ) = 0.4 + 0.1 + 0.05 = 0.55

(iv) P ( 2 < X < 5 ) = P ( X = 3) + P ( X = 4 ) = 0.4 + 0.1 = 0.5

(v) E ( X ) = 1( 0.2 ) + 2 ( 0.25 ) + 3 ( 0.4 ) + 4 ( 0.1) + 5 ( 0.05 ) = 0.2 + 0.5 + 1.2 + 0.4 + 0.25 = 2.55 (vi) Mode is 3 Question # 2 Var ( X ) = E ( X 2 ) −  E ( X ) 

2

S .D = Var

,

(a) E ( X ) = 1× 0.1 + 2 × 0.3 + 3 × 0.2 + 4 × 0.3 + 5 × 0.1 = 3 (b) Now E ( X

) = 1 × 0.1 + 2 × 0.3 + 3 × 0.2 + 4 × 0.3 + 5 Var ( X ) = E ( X ) −  E ( X )  = 10.4 − 3 = 1.4 2

2

2

2

2

2

2

2

× 0.1 = 10.4

2

(c) S .D. = Var ( X ) = 1.4 = 1.1832 ≈ 1.2 Question # 3 P.D.F. = Probability Density Function P ( X = x ) = kx, x = 8,9,10

P ( X = 8 ) = 8k , P ( X = 9 ) = 9k , P ( X = 10 ) = 10k ∴ 8k + 9k + 10k = 1 27k = 1 1 k= 27 Question # 4 2 (a) P ( Y = y ) = my , y = 0,1, 2,3, 4

P ( Y = 0 ) = 02 m = 0 P ( Y = 1) = 12 m = m

P ( Y = 2 ) = 2 2 m = 4m P ( Y = 3) = 32 m = 9m P ( Y = 4 ) = 42 = 16m m + 4m + 9m + 16m = 1 1 30m = 1 ∴m = 30 (b) y P(Y = y) E ( Y ) = 1×

0 1 0 1 30

2 4 30

3 9 30

4 16 30

1 4 9 16 1 8 27 64 100 1 + 2 × + 3× + 4 × = + + + = =3 30 30 30 30 30 30 30 30 30 3

1

Question # 5 x 0 1 2 3 4 5 6

(a)

P(X) 0.32 0.35 0.18 0.08 0.04 0.02 0.01

0 0.35 0.36 0.24 0.16 0.10 0.06

E(X) = 1.27 E ( X 2 ) = 12 × 0.35 + 22 × 0.18 + 32 × 0.08 + 42 × 0.04 + 52 × 0.02 + 62 × 0.01 = 0.35 + 0.72 + 0.72 + 0.64 + 0.5 + 0.36 = 3.29

Var ( X ) = E ( X 2 ) −  E ( X )  = 3.29 − ( 1.27 ) = 1.6771 2

2

(b) Standard Deviation = 1.6771 = 1.295 Question # 6 Y P(Y = y)

10 P1

20 P2

10 P1 + 20 P2 = 16 ..................equ (1) P1 + P2 = 1......................equ (2) × 10 From equation (2) 10 P1 + 10 P2 = 10...equ (3) [ Subtracting equ (3) from equ (1)] 10 P2 = 6 P2 = 0.6 ∴ P1 = 1 − 0.6 = 0.4 Y P(Y = y)

10 0.4

20 0.6

Question # 7 n = 3 , p = 0.05 , q = 0.95 0 3 P ( X = 0 ) = 3C0 p 0 q3−0 = 3C0 ( 0.05 ) ( 0.95 ) = 0.857 (a) (i) (ii)

P ( X ≥ 1) = P ( X = 1) + P ( X = 2 ) + P( X = 3) = 0.135 + 0.007 + 0.000 = 0.142

OR = 1 −  P ( X = 0 )  =1- 0.857 = 0.143 (iii) (b)

P ( X > 1) = P ( X = 2 ) + P ( X = 3) = 0.007 + 0.000 = 0.007

E ( X ) = np = 3 × 0.05 = 0.15 Var ( X ) = npq = 3 × 0.05 × 0.95 = 0.1425 Standard Deviation = npq = 0.1425 ; 0.3775

Question # 8 n x n−x N.B. P ( X = x ) = C x p q n = 10; p = 0.2 ; q = 0.8 0 10 −0 a) P ( X = 0 ) = 10C0 p 0 q10 −0 = 10C0 ( 0.2 ) ( 0.8 ) = 0.107 b) (Exactly) P( X = 2 ) = 10C2 ( 0.2 ) c) d)

( 0.8) = 0.302 (At least) P ( X ≥ 2 ) = P ( X = 2 ) + P ( X = 3) .... + P ( X = 10 ) OR P ( X ≥ 2 ) = 1 − [ P ( X = 0) + P ( X = 1) = 1 − [ 0.107 + 0.268] = 0.625 2

10 −2

P ( X ≤ 2) = P ( X = 0 ) + P ( X = 1) + P ( X = 2 ) = 0.107 + 0.268 + 0.302 = 0.677 2

Question # 9 n=5 (a) (i) (ii) (iii)

p = 0.70 q = 0.30 P ( X = 5 ) = 0.168

P ( X ≥ 3) = P ( X = 3) + P ( X = 4 ) + P( X = 5) = 0.309 + 0.360 + 0.168 = 0.837 P ( X < 2 ) = P ( X = 0 ) + P ( X = 1) = 0.002 + 0.028 = 0.030

(b) E ( X ) = np = 5 × 0.7 = 3.5, Var ( X ) = npq = 5 × 0.7 × 0.3 = 1.05, ∴ S. D. = npq = 1.05 ; 1.025 Question # 10 (Use Formula)

(i)

n = 10 p = 0.02 q = 0.98 0 10 10 0 10 −0 P ( X = 0 ) = C0 p q = 10C0 ( 0.02 ) ( 0.98 ) = 0.817

(ii)

P ( X = 1) = 10C1 p1q10 −1 = 10C1 ( 0.02 ) ( 0.98 ) = 0.167

(iii)

1

9

P ( X ≥ 2 ) = 1 − [ P ( X = 0) + P ( X = 1) = 1 − [ 0.817 + 0.167 ] = 0.984

Question # 11 λ =3 a) P ( X < 3) = P ( X = 0 ) + P ( X = 1) + P ( X = 2 ) = 0.0498 + 0.1494 + 0.2240 = 0.4232 b) P ( X = 3) = 0.2240

c) P ( X ≥ 3) = 1 −  P ( X = 0 ) + P ( X = 1) + P ( X = 2 )  = 1 – 0.4232 = 0.5768

d) P ( X > 3) = 1 −  P ( X = 0 ) + P ( X = 1) + P ( X = 2 ) + P ( X = 3)  = 1 – 0.6472 = 0.3528 Question # 12 λ =9 a) P ( X = 7 ) = 0.1171 b) P ( X < 5 ) = P ( X = 0 ) + P ( X = 1) + P ( X = 2 ) + P ( X = 3) + P ( X = 4 ) = 0.0001 + 0.0011 + 0.0050 + 0.0150 + 0.0337 = 0.0549 Question # 13 λ =6 a) P ( X < 5 ) = P ( X = 0 ) + P ( X = 1) + P ( X = 2 ) + P ( X = 3) + P ( X = 4 ) = 0.2851 b) P ( X = 5 ) = 0.1606

c) P ( X ≥ 5 ) = 1 −  P ( X = 0 ) + P ( X = 1) .... + P ( X = 4 )  = 0.7149 d) P ( X = 4 ) + P ( X = 5 ) = 0.1339 + 0.1606 = 0.2945

Question # 14 λ=2 a) P ( X > 3) = 1 – [ P ( X = 0 ) + P ( X = 1) + P ( X = 2 ) + P ( X = 3) ] = 1 – [0.1353 + 0.2707 + 0.2707 + 0.1804 = 1 – 0.8571 = 0.1429 b) λ = 4 [Fortnight (2 weeks)] P [X > 2] = P ( X = 0 ) + P ( X = 1) + P ( X = 2 ) = 1 – [0.0183 + 0.0733 + 0.1465 = 0.7619 2 c) Given day (Assume 5 day week) λ = = 0.4 5 P (X = 0) = 0.6703 Question # 15

λ = 2 for 100m ∴ for 200m λ = 4 a) P ( X > 3) = 1 – [ P ( X = 0 ) + P ( X = 1) + P ( X = 2 ) + P ( X = 3) ] = 1 – 0.4335 = 0.5665 b) 50m ⇒ λ = 1 P ( X = 2 ) = 0.1839 3