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Homework #4 Solution • Daily amount of raining in winter at Gaza is by ml cube, A random variable X have a continuous uniform distribution with A = 100 ml cube , B = 170 ml cube . • Find the probability that a given day the amount of raining is: Homework #4 Solution • At most 110 ml cube. A= 100 , B = 170 At most 110 110 100 P( x 110) 0.143 170 100 Homework #4 Solution 1. More than 120 ml cube but less than 160 ml cube. 160 120 P(120 X 160) 0.571 170 100 Homework #4 Solution • At least 145 170 140 P( X 140) 0.429 170 100 Homework #4 Solution • A given normal distribution finds the value of k such: 1. P( z < k) = 0.0078 From table k = - 2.42 2. P( z > k) = 0.0099. P(z > k ) = 1-.0099=0.9901 Then k = 2.33 Homework #4 Solution • A given normal distribution finds the value of k such: 3. P( -0.88 < z <k ) = 0.8106 Area left of z = - 0.88= 0.1894 Then 0.1894 + 0.8106 = 1 = k K = 3.49 4. t15,0.02 = k k = 2.249 Homework #4 Solution • 9 basketball players through balls into a ring if the probability to score is 0.4 find: • At least 4 players are scored n= 9 , p = 0.4 , q = 0.6 9 P(0) (0.4) 0 (0.6) 9 0.01 0 Homework #4 Solution 9 1 8 P(1) ( 0 . 4 ) ( 0 . 6 ) 0.06 1 9 2 7 P(2) ( 0 . 4 ) ( 0 . 6 ) 0.161 2 9 P(3) (0.4) 3 (0.6) 6 0.251 3 At least 4 players are scored = 1-P(0)+P(1)+P(2)+P(3) =1- (0.01 + .06 + 0.161 + 0.251) = 0.518 Homework #4 Solution • At least 5 players failed to score p=0.6, q=0.4, At least 5 players failed to score = 1 – (P(0)+P(1)+P(2)+P(3)+P(4) 9 P(0) (0.6) 0 (0.4) 9 0.0.000262 0 9 P(1) (0.6)1 (0.4) 8 0.00354 1 9 2 7 P(2) (0.6) (0.4) 0.0122 2 Homework #4 Solution 9 3 6 P(3) ( 0 . 6 ) ( 0 . 4 ) 0.0743 3 9 4 5 P(4) 4 (0.6) (0.4) 0.167 At least 5 players failed to score = 1 – (P(0)+P(1)+P(2)+P(3)+P(4) = [ 1 – (0.000262 + 0.00354 + 0.0122 + 0.0743 + 0.167)]=0.7427 • Exactly 6 players are scored 9 P(6) (0.4) 6 (0.6) 3 0.0743 6 Homework #4 Solution • An experiment was run to test many of thermometers to test the freezing point of water. Assume that the mean reading is 0 c and the standard deviation is 1.1 c and assume that the readings are normal distribution. • If one thermometer selected randomly find: Homework #4 Solution 1. The reading of freezing point is less than 1.58 c. 1.58 0 z 1.436 1.1 Then the area is 0.9251, which is the probability of reading less than 1.58. Homework #4 Solution • • The reading of freezing point is above 1.19c. Above -1.19 z 1.19 0 1.08 1.1 Homework #4 Solution • The reading of freezing point is between 1.6c and -1.2c. 1.2 0 1.09 1.1 1.6 0 z (1.6) 1.45 1.1 z (1.2) Area = 0.9265 - 0.1379 = 0.7886 Homework #4 Solution • A fabric manufacturer believes that the proportion of orders for row material arriving late is P=0.62. • If random sample of 50 orders show that 4 or fewer arrived late, the hypothesis that P=0.62 should be rejected in favor of the alternative p < 0.62, use binomial distribution: 1. Find the probability of committing a type I error. Homework #4 Solution P( x 4 | p 0.62) p(0) p(1) p(2) p(3) p(4) | p 0.62 50 p (0) (0.62 0 )(0.38 50 ) 9.8 10 22 0 50 p (1) (0.621 )(0.38 49 ) 7.96 10 20 1 50 p (2) (0.62 2 )(0.38 48 ) 3.18 10 18 2 50 p (3) (0.62 3 )(0.38 47 ) 8.3 10 17 3 50 p (4) (0.62 4 )(0.38 46 ) 1.6 10 15 4 p(1) p(2) p(3) p(4) Homework #4 Solution • Find the probability of committing type II error for alternatives p=0.29, p=0.38, p=0.48 50 50 50 0 50 1 49 2 48 (0.29 )(0.71 ) (0.29 )(0.71 ) (0.29 )(0.71 ) 0 1 2 1 1 50 50 3 47 4 46 (0.29 )(0.71 ) (0.29 )(0.71 ) 3 4 Homework #4 Solution 50 50 50 0 50 1 49 2 48 (0.38 )(0.62 ) (0.38 )(0.62 ) (0.38 )(0.62 ) 0 1 2 2 1 50 50 3 47 4 46 (0.38 )(0.62 ) (0.38 )(0.62) 3 4 50 50 50 0 50 1 49 2 48 (0.48 )(0.52 ) (0.48 )(0.52 ) (0.48 )(0.52 ) 0 1 2 3 1 50 50 3 47 4 46 (0.48 )(0.52 ) (0.48 )(0.52 ) 3 4 Homework #4 Solution • The average height of girls in a class at past year was 162.5 cm with standard deviation of 6.9cm. If random sample of 20 girls were selected test the claim that the average height are 165.5 cm at alpha = 0.01. H0: µ=165.5 Ha: µ≠165.5 Homework #4 Solution • Calculated t t n( X ) 20 (165.5 162.5 1.9544 6.9 Tabulated t t 0.01 ,19 2 2.861 Calculated t < tabulated t then We cannot reject the H0 Homework #4 Solution • State the regression equation with suitable hypothesis and calculate errors for : y 10 37 40 44 38 47 60 x 12 23 24 22 29 35 38 Homework #4 Solution Sum y 10 37 40 44 38 47 60 276 x 12 23 24 22 29 35 38 183 X^2 144 529 576 484 841 1225 1444 5243 xy 120 851 960 968 1102 1645 2280 7926 Estimate (y) 17.528 34.567 36.116 33.018 43.861 53.155 57.802 errors -7.528 2.433 3.884 10.982 -5.861 -6.155 2.198 Homework #4 Solution • N = 7; ; (∑X)2 = 5243; (∑X) (∑Y) = 7926 y 39.43, x 26.14 S xy X Y 183 276 XY 7926 710.57 n 7 Homework #4 Solution S xx X 2 X 2 n (183 2 ) 5243 458.857 7 S XY 710.57 b1 1.549 S XX 458.857 b0 y b1 x 39.43 (1.549 26.14 1.06 Homework #4 Solution • For b0 H 0 : b0 0 H a : b0 0 For b1 H 0 : b1 0 H a : b1 0 b0 1.06 0 Then it’s affected on model Homework #4 Solution b1 1.549 0 Then it’s affected on model Then the model is: yˆ 1.06 1.549 x ˆ Errors Y Y