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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:
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Homework #4 Solution At most 110 ml cube. A= 100, B = 170 At most 110
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Homework #4 Solution 1.More than 120 ml cube but less than 160 ml cube.
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Homework #4 Solution At least 145
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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
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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. t 15,0.02 = k k = 2.249
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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
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Homework #4 Solution 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
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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)
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Homework #4 Solution 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
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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:
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Homework #4 Solution 1.The reading of freezing point is less than 1.58 c. Then the area is 0.9251, which is the probability of reading less than 1.58.
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Homework #4 Solution The reading of freezing point is above - 1.19c. Above -1.19
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Homework #4 Solution The reading of freezing point is between 1.6c and -1.2c. Area = 0.9265 - 0.1379 = 0.7886
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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.
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Homework #4 Solution
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Find the probability of committing type II error for alternatives p=0.29, p=0.38, p=0.48
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Homework #4 Solution
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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. H 0 : µ=165.5 H a : µ≠165.5
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Homework #4 Solution Calculated t Tabulated t Calculated t < tabulated t then We cannot reject the H 0
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Homework #4 Solution State the regression equation with suitable hypothesis and calculate errors for : y10374044384760 x12232422293538
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Homework #4 Solution
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N = 7; (∑X)2 = 5243; (∑X) (∑Y) = 7926 ;
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Homework #4 Solution
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For b 0 For b 1 Then it’s affected on model
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Homework #4 Solution Then it’s affected on model Then the model is:
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