mat2378-practice-mc-2

mat2378-practice-mc-2 - Multiple choice set 2 INFO practice...

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Unformatted text preview: Multiple choice set 2 INFO: practice multiple choice questions related to chapters 3, 4, 5 and assignment 3. Q1. A factory employs several thousand workers, of whom 30% are from non-English speaking background. If 15 members of the union executive committee were chosen from the workers at random, evaluate the probability that exactly 3 members of the committee are non-English background people. (a) 0.17 (b) 0.83 (c) 0.98 (d) 0.51 (e) none of the preceding Solution to Q1: Y - number with non-English background. Y ∼ B (15, 0.3), to compute P (Y = 3) = 0.17 Q2. On average, one person in a hundred will carry a certain mutant gene. 60 people are tested. What is the probability that 5 or more of these people will be found to carry the gene? (choose the closest answer) (a) 1.00000 (b) 0.99965 (c) 0.00035 (d) 0.01 (e) none of the preceding The MINITAB output will help you to solve this problem: Binomial with n = 60 and p = 0.01 x 0 1 2 3 4 P( X = x ) 0.547157 0.331610 0.098813 0.019297 0.002778 Solution to Q2: success= a person carries the gene, thus P (success) = p = 0.01. If X - number of person which carry the gene; X ∼ B(60, 0.01). To compute: P (X ≥ 5) = 1 − P (X ≤ 4) = 1 − 0.99965. Q3. (5.5) In a certain forest, 25% of the white pine tree are infected with blister rust. Suppose a sample of four white pine trees is to be chosen. Let p be the sample proportion of infected trees. The probability that p = 0.25 ˆ ˆ is (a) 0.3164 (b) 0.2109 (c) 0.25 (d) 0.4219 (e) none of the preceding Q4. If X ∼ N (0, 4) the value of P (|X | ≥ 2.2) is (a) 0.2321 (b) 0.8438 Solution to Q4: (c) 0.2526 (d) 0.2714 (e) 0.7286 P (|X | ≥ 2.2) = = = 1 − P (|X | ≤ 2.2) = 1 − P (−2.2 ≤ X ≤ 2.2) = ￿ ￿ −2.2 − 0 X −0 2.2 − 0 √ 1−P ≤√ ≤√ = 4 4 4 1 − (P (Z < 1.1) − P (Z < −1.1)) = 1 − (0.8643 − 0.1357) = 0.2714. Q5. If X ∼ N (10, 1), the value of k such that P (X ≤ k ) = 0.7019 is closest to (a) 0.59 (b) 0.30 (c) 0.53 (d) Solution to Q5: P (X ≤ k ) = P Thus, k − 10 = 0.53 and k = 10.53. ￿ X − 10 k − 10 ≤ 1 1 ￿ 10.53 (e) none of the preceding = P (Z ≤ k − 10) = 0.701944 Q6. Suppose that X1 ∼ N (3, 4) and X2 ∼ N (3, 45). Given that X1 and X2 are independent random variables, P (X1 + X2 > 9.5) is (use the normal table) (a) 0.3085 (b) 0.6915 (c) 0.5279 (d) 0.4271 (e) none of the preceding Solution to Q6: X1 + X2 ∼ N (3 + 3, 4 + 45) = N (6, 49). Thus, let Y = X1 + X2 ￿ ￿ Y −6 9.5 − 6 P (X1 + X2 > 9.5) = P (Y > 9.5) = P > = P (Z > 0.5) = 1 − 0.691462 7 7 Q7. Suppose that the weight in pounds of a North American adult can be represented by a normal random variable with mean 150 lbs. and variance 900 lb2 . An elevator containing a sign “Maximum 12 people” can safely carry 2000 lbs. The probability that 12 people will not overload the elevator is closest to (a) 0.9729 (b) 0.4501 (c) 0.0271 (d) 0.0001 (e) 1.3 Solution to Q7: Let S - total weight of 12 people. S ∼ N (12 × 150, 12 × 900) = N (1800, 10800). ￿ ￿ S − 1800 2000 − 1800 P (S > 2000) = P √ >√ = 0.0271 10800 10800 is the probability that they will overload. Thus, answer 0.9729. Q8. Let X1 , . . . , X50 be a random sample from a populaiton with mean 1 and variance 1. The approximate proba50 ￿ bility P (48 ≤ S ≤ 52), where S = Xi , is closest to i=1 (a) 0.6368 (b) 0.4534 (c) 0.2227 (d) 0.9988 (e) 0.5000 Solution to Q8: We have E(S ) = 50, Var(S ) = 50. So, according to CLT ￿ ￿ 48 − 50 S − 50 52 − 50 √ P (48 ≤ S ≤ 52) = P ≤√ ≤√ = P (−0.2828 < Z < 0.2828) ≈ 0.2227 50 50 50 Q9. The heights of a certain population of corn plants follow a distribution with mean 145 cm and standard deviation ¯ 22 cm. If Y represents the mean height of a random sample of 100 plants, then the approximate value of ¯ ≤ 140) is P (Y (a) 0.0116 (b) (c) (d) (e) 0.1423 0.3336 2 none of the preceding Q10. Which of the following set of figures describes data coming from the same normal population? (a) (b) (c) (d) Solutions to multiple choice questions: Q1 −→ a Q2 −→ c Q3 −→ d Q4 −→ d Q5 −→ d Q6 −→ a Q7 −→ a Q8 −→ c Q9 −→ a Q10 −→ b ...
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This note was uploaded on 01/13/2010 for the course MATH MAT2378 taught by Professor Kulik during the Spring '09 term at University of Ottawa.

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