Statistics Ii Worksheet # 2 (solutions)(2)

UNIVERSITY OF TECHNOLOGY, JAMAICA STATISTICS II (STA 2004) WORKSHEET # 2 (SOLUTIONS) Question # 1 Using Normal Distribution tables, evaluate: i. P ( Z > 1.27) = 0.3980 = 0.5000 − 0.3980 = 0.1020 ii.

P ( Z > -2.71) = 0.4966 + 0.5 = 0.9966

iii.

P ( Z < -0.72) = 0.5 – 0.2642 = 0.2358

iv.

P ( Z < 2.09) = 0.5 + 0.4817 = 0.9817

v.

P ( 0.87 < Z < 2.35) = 0.1828

vi.

P ( - 1.89 < Z < 0.6) = 0.6963

vii.

P ( - 2.04 < Z < 1.46) = 0.4793 + 0.4279 = 0.9072

viii.

P ( -2.44 < Z < -0.64) = 0.2538

Question # 2 Given µ = 368 and σ = 97 i. P ( 115 < X) Standard Normal Variable X =µ Z= σ  115 − 368  P < Z  = P ( −2.608 < Z ) = 0.4955 + 0.5 = 0.9955 97  

ii.

iii.

  594 − 368   P ( X > 594 ) = P  Z >   97      226   = P Z >    = P ( Z > 2.329 ) = 0.5 − 0.4901 = 0.0099  97      229 − 368   P ( 229 < X ) = P  
Question # 2 continued iv.

v.

  428 − 368   P ( X > 428 ) = P  
vi.

609 − 368   458 − 368 P ( 458 < X < 609 ) = P 
vii.

‘a’ if P ( X < a ) = 0.1466 µ = 368 & σ = 97 X −µ Z= σ  X − µ a − 368  P <  = 0.1466 97   σ a − 368   Equation 1 → P  Z <  = 0.1466 97   Equation 2 → P ( Z < −1.05 ) = 0.1466 Comparing equation 1 & equation 2 a − 368 = −1.05 97 a = ( −1.05 × 97 ) + 368 = 368 − 101.85 a = 266.15 ∴ P ( X < 266.15 ) = 0.1466

viii.

P ( X > b ) = 0.3745  X − µ b − 368  P >  = 0.3745 97   σ b − 368   Equation 1 → P  Z >  = 0.3745 97   Equation 2 → P ( Z > 0.32 ) = 0.3745 Comparing equations 1 & 2 b − 368 = 0.32 97 b = ( 97 × 0.32 ) + 368 b = 399.09 ∴ P ( Z > 399.09 = 0.3745 )

Question # 3 Page 2

i.

P ( 1820 < X < 1970 )  1820 − 1820 X − µ 1970 − 1820  = P < <  120 σ 120   150   = P0 < Z <  = P ( 0 < Z < 1.25 ) = 0.3944 = 39.44% 120  

ii.

P ( X ≥ 1970 ) 1970 − 1820   = PZ ≥  120   = P ( Z ≥ 1.25 ) = 0.1056 = 10.56%

iii.

P ( X ≤ 1600 ) 1600 − 1820   = PZ ≤  120   220   = PZ ≤ −  = P ( Z ≤ −1.83) = 0.0336 = 3.36% 120  

Question # 4

µ = 32 hours i.

σ = 2 hours  32 − 32 X − µ 34 − 32  P ( 32 < x < 34 ) = P  < <  σ 2   2 = P ( 0 < Z < 1) = 0.3413 = 34.13%

ii.

iii.

28.7 − 32   P ( X ≤ 28.7 ) = P  Z ≤  2   3.3   = PZ ≤ −  = P ( Z ≤ −1.65 ) 2   = 0.0495 = 4.95% P ( X ≥ a ) = 0.05  X −µ a−µ  P ≥ = 0.05 σ   σ a − 32   Equation 1 → P  Z ≥  = 0.05 2   Equation 2 → P ( Z ≥ 1.65 ) = 0.05 From equations 1 & 2 a − 32 = 1.65 2 a = ( 1.65 × 2 ) + 32 a = 35.3 hours

Question # 5

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µ = 20 seconds σ = 8 seconds P ( X > 30 ) i. 30 − 20  10    = PZ >  = P  Z >  = P ( Z > 1.25 ) 8  8   = 0.5 − 0.3944 = 0.1056 ii.

P ( X < 15 ) 15 − 20   = PZ <  = P ( Z < −0.625 ) 8   = 0.5 − 0.2357 = 0.2643

iii.

P ( 15 < X < 25 ) 25 − 20   15 − 20 = P
iv.

P ( X < a ) = 0.4 a − 20   Equation 1 → P  Z <  = 0.4 8   Equation 2 → P ( Z < −0.25 ) = 0.4 From equations 1 & 2 a − 20 = −0.25 8 a = ( −0.25 × 8 ) + 20 = 18 seconds

Question # 6 i.

ii. iii.

iv.

5200 − 3462   P ( X > 5200 ) = P  Z >  = P ( Z > 1.62 ) 1070   = 0.5 − 0.4474 = 0.0526 4000 − 3462   P ( X < 4000 ) = P  Z <  = P ( Z < 0.5028 ) 1070   2900 − 3462 2900 − 3462   P ( 2900 < X < 5500 ) = P  a ) = 0.85

a − 3462   Equation 1 → P  Z >  = 0.85 1070   Equation 2 → P ( Z > −1.04 ) = 0.85 a − 3462 = −1.04 1070 a = ( −1.04 ×1070 ) + 3462 = −1112.8 + 3462 = 2349.2

∴

v. vi.

P ( X > a ) = 0.20 P ( X < a ) = 0.75

4360.8 hours 4189.6 hours

BINOMIAL APPROXIMATION Page 4

Question # 7 The Normal Approximation to the Binomial Distribution p = 0.1 & q = 0.9 µ = E ( X ) = np = (500) (0.1) = 50 Var ( X ) = npq

= ( 500 ) ( 0.1) ( 0.9 ) = ( 50 ) ( 0.9 ) = 45

σ = 45 c .c P ( X > 60 )  → P ( X > 60.5 )

 X − µ 60.5 − 50  = P >  = P ( Z > 1.57 ) = 0.5 − 0.4418 = 0.0582 45   σ Question # 8 n = 400 Var ( X ) = npq ∴σ = Var ( X ) p = 0.10 µ = np q = 0.90 µ = 400 × 0.1 = 40 Var ( X ) = npq = ( 400 ) ( 0.1) ( 0.9 ) = 36

σ = 36 = 6 c .c . P ( X ≥ 35 )  → P ( X > 34.5 ) i. 34.5 − 40   = PZ >  = P ( Z > −0.92 ) = 0.3212 + 0.5 = 0.8212* 6   ii.

c .c . P ( 40 ≤ X ≤ 50 )  → P ( 39.5 < X < 50.5 )

50.5 − 40   39.5 − 40 = P
c .c . P ( 34 < X < 48 )  → P ( 34.5 < X < 47.5 )

47.5 − 40   34.5 − 40 = P
Question # 9

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n = 45 p = 0.28 q = 0.72 µ = E ( X ) = np = 45 × 0.28 = 12.6 Var ( X ) = npq = ( 12.6 ) ( 0.72 ) = 9.072

σ = 9.072 = 3.01 c .c . P ( X > 15 )  → P ( X > 15.5 ) i. 15.5 − 12.6   = PZ >  = P ( Z > 0.96 ) = 0.5 − 0.3315 = 0.1685 3.01   c .c . P ( 10 < X < 20 )  → P ( 10.5 < X < 19.5 ) ii. 19.5 − 12.6   10.5 − 12.6 = P
ii.

c .c . P ( X > 30 )  → P ( X > 30.5 )

30.5 − 25   = PZ >  = P ( Z > 1.1) = 0.5 − 0.3643 = 0.1357 5   P ( 23 ≤ X ≤ 29 ) → P ( 22.5 < X < 29.5 ) 29.5 − 25   22.5 − 25 = P
Question # 11

λ = 36 i.

Var ( X ) = 36

σ =6

P ( X < 30 ) → P ( X < 29.5 ) c .c .

29.5 − 36   = PZ <  = P ( Z < −1.08 ) = 0.5 − 0.3599 = 0.1401 6   ii.

c .c . P ( X > 40 )  → P ( X > 40.5 )

40.5 − 36   = PZ >  = P ( Z > 0.751) = 0.5 − 0.2734 = 0.2266 6   iii.

c .c . P ( 30 ≤ X ≤ 40 )  → P ( 29.5 < X < 40.5 )

40.5 − 36   29.5 − 36 = P
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