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QMT202/SET2

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QMT202/SET2

1. 1. PART B QUESTION 1 (a) X ~ Bin (9, 0.45) 9 C ( 0.45 ) .(0.55)5 4 i) P( X = 4 ) = 4 = 0.26 ii) P( X ≥7 ) = P ( X = 7 ) +P ( X = 8 ) +P ( X = 9 ) 9 9 9 C ( 0.45) .(0.55)2 + C ( 0.45) .(0.55)1 + C ( 0.45) .(0.55)0 7 8 9 = 7 8 9 = 0.0498 iii) P (5 ≤ X ≤ 7 ) = P (X = 5 ) + P ( X = 6 ) + P ( X = 7 ) 9 9 9 C ( 0.45) .(0.55)4 + C ( 0.45) .(0.55)3 + C ( 0.45) .(0.55)2 5 6 7 = 5 6 7 = 0.3695 QUESTION 1 (b) X ~ N (25, 6 2 ) 16 − 25 i) P (X < 16) = P (z < ) 6 = P ( z < - 1.5 ) = 0.5 – 0.4332 = 0.0668 22 − 25 29 − 25 ii) P ( 22 ≤ X ≤ 29 ) = P ( ≤z≤ ) 6 6 = P ( - 0.5 ≤ z ≤ 0.67 ) = 0.2486 + 0.1915 = 0.4401 iii) P ( X ≥ a ) =0.1075 P ( X < a ) =0.8925  x − 75 a − 75  P <  = 0.8925  10 10   a − 75  P z <  = 0.8925  10  From the table P ( z < 1.24 ) = 0.8925
2. 2. a − 75 = 1.24 10 a = 87.4 QUESTION 2 (a)  3.5  X ~N  40,   49  3.5 i) σX = = 0.5 49 41 − 40 P( X ≥ 41 ) = P( z ≥ ) 0.5 = P(z ≥ 2) = 0.5 – 0.4772 = 0.0228 ii) P(40.5 < X < 40.9) 40.5 − 41 40.9 − 41 = P( <z< ) 0.5 0.5 = P(-1 < z < -0.2) = 0.3413 – 0.0793 = 0.262 QUESTION 2 (b) 30 n = 150, p = 150 30 i) p= ˆ = 0.2 150 (0.2)(0.8) σp = ˆ = 0.0327 150 margin of error = ± 1.96σ p ˆ
3. 3. = ±1.96(0.0327) = ±0.0641 ii) From table, the z-value for 95% CI = 1.96 The 95% Confidence Interval for population mean, = p ± zσ p ˆ ˆ = 0.2 ± 1.96(0.0327) = 0.072 ± 0.0641 = (0.1359, 0.2641) 95% of confidence interval for the percentage of cakes being effect with fungus is lies between 0.1359 to 0.2641 iii) From table, the z-value for 90% CI = 1.65 E = 0.05 2  zs  n = p.q.   E 2  1.65  = 0.2 × 0.8 ×    0.05  = 174.24 The sample size needed is 174 . QUESTION 3(a) σ = 1.75 n = 35 , x = 7.32 i. H0 : µ ≥ 8 H1 : µ < 8 ii. Since n is large, 35 >30, we use normal distribution to make the test. x−µ 7.32 − 8 = Test statistic, z = σ 1.75 = - 2.2988 n 35 iii. At α = 0.05 , we reject H0 if z < -1.65 Since - 1.65 > -2.29 , we reject H0 We conclude that there is no evidence that the effective rate of broadband networking for High-Speed Company production of electronic is at least 8.
4. 4. QUESTION 3(b) 24 n= 505 , p= ˆ = 0.0475 505 24 i. p= ˆ = 0.0475 505 ii. H0 : p = 0.35 H1 : p ≠ 0.35 iii. Since np > 5 and nq > 5 , we use normal distribution to make the test . p − p 0.0475 − 0.35 ˆ = Test statistic , z = pq 0.35 × 0.65 = -14.25 n 505 At α = 0.01 , we reject H0 if z < -2.58 or z > 2.58 Since -2.58 > -14.25 , we reject H0. We conclude that the proportion of the safety rate of the building is not equal 0.35 QUESTION 4(a) i. Independent variable: Driving Experience (years) ii. Scatter diagram Label x -axis Label y -axis Title Diagram iii.
5. 5. ∑ x = 1396 2 ∑ x = 90 n=8 ∑ y = 29642 2 ∑ y = 474 ∑ xy = 4739 n∑ xy − ∑ x ∑ y r= n x2 − ( x ) 2  n y 2 − (  ∑ ∑  ∑ ∑ y)  2     8 ( 4739 ) − 90 ( 474 ) = 8 ( 1396 ) − 902  8 ( 29642 ) − 4742     = −0.7679 There is a negative correlation between the variables. QUESTION 4(b) i. There is a strong positive correlation between cost and age. ii. n∑ xy − ∑ x ∑ y b= n∑ x 2 − ( ∑ x ) 2 6 ( 3100 ) − 49 ( 325 ) = 6 ( 481) − 492 = 5.515 a = y − bx = 54.17 − 5.515 ( 8.17 ) = 9.112 Regression line: y = 9.112 + 5.515 x iii. 0.96862 = 0.9382 . 93.82% of the total variation in Cost is explained by Age and another 6.18% is explained by another factors. iv. y = 9.112 + 5.515 x ˆ = 9.112 + 5.515 ( 16 ) = 97.35