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5.5 Solve the following problems by using the binomial formula. If n = 4 and p = .10, find P(x = 3). If n = 7 and p = .80, find P(x = 4). If n = 10 and p = .60, find P(x ≥ 7). If n = 12 and p = 45, find D a. b. c. d.

5.5 Solve the following problems by using the binomial formula. If n = 4 and p = .10, find P(x = 3). If n = 7 and p = .80, find P(x = 4). If n = 10 and p = .60, find P(x ≥ 7). If n = 12 and p = 45, find D a. b. c. d.

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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DISTRIBUTIONS stretches toward the higher values of x. The mean of the distribution n = 8 and p = .80 is 6.4, which results in the highest probabilities being near x = 6 and x = 7. Thus the peak of the distribution is nearer to 8 than to 0 and the distribution stretches back toward x = 0. In any binomial distribution the largest x value that can occur is n and the smallest value is zero. Thus the graph of any binomial distribution is constrained by zero and n. If the p value of the distribution is not .50, this constraint will result in the graph "piling up" at one end and being skewed at the other end. A manufacturing company produces 10,000 plastic mugs per week. This company supplies mugs to another company, which packages the mugs as part of picnic sets. The second company randomly samples 10 mugs sent from the supplier. If two or fewer of the sampled mugs are defective, the second company accepts the lot. What is the probability that the lot will be accepted if the mug manufacturing company actually is producing mugs that are 10% defective? 20% defective? 30% defective? 40% defective? Solution In this series of binomial problems, n = 10, x ≤ 2, and p ranges from .10 to .40. From Table A.2-and cumulating the values-we have the following probability of x ≤ 2 for each p value and the expected value (= n. p). Expected Number Lot Accepted Pix ≤ 2) of Defects (μ) Р .10 .930 1.0 .20 .677 2.0 .30 .382 3.0 .40 .167 4.0 These values indicate that if the manufacturing company is producing 10% defective mugs, the probability is relatively high (.930) that the lot will be accepted by chance. For higher values of p, the probability of lot acceptance by chance decreases. In addition, as p increases, the expected value moves away from the acceptable values, x≤ 2. This move reduces the chances of lot acceptance. 5.5 Solve the following problems by using the binomial formula. a. If n = 4 and p = .10, find P(x = 3). b. If n = 7 and p = .80, find P(x = 4). C. If n = 10 and p = .60, find P(x ≥ 7). d. If n = 12 and p = .45, find P(5 ≤ x ≤ 7). Solve the following problems by using the binomial tables (Table A.2). a. If n = 20 and p = .50, find P(x = 12). b. If n = 20 and p = .30, find P(x > 8). C. If n = 20 and p = .70, find P(x < 12). d. If n = 20 and p = 90, find 5.6
Transcribed Image Text:DISTRIBUTIONS stretches toward the higher values of x. The mean of the distribution n = 8 and p = .80 is 6.4, which results in the highest probabilities being near x = 6 and x = 7. Thus the peak of the distribution is nearer to 8 than to 0 and the distribution stretches back toward x = 0. In any binomial distribution the largest x value that can occur is n and the smallest value is zero. Thus the graph of any binomial distribution is constrained by zero and n. If the p value of the distribution is not .50, this constraint will result in the graph "piling up" at one end and being skewed at the other end. A manufacturing company produces 10,000 plastic mugs per week. This company supplies mugs to another company, which packages the mugs as part of picnic sets. The second company randomly samples 10 mugs sent from the supplier. If two or fewer of the sampled mugs are defective, the second company accepts the lot. What is the probability that the lot will be accepted if the mug manufacturing company actually is producing mugs that are 10% defective? 20% defective? 30% defective? 40% defective? Solution In this series of binomial problems, n = 10, x ≤ 2, and p ranges from .10 to .40. From Table A.2-and cumulating the values-we have the following probability of x ≤ 2 for each p value and the expected value (= n. p). Expected Number Lot Accepted Pix ≤ 2) of Defects (μ) Р .10 .930 1.0 .20 .677 2.0 .30 .382 3.0 .40 .167 4.0 These values indicate that if the manufacturing company is producing 10% defective mugs, the probability is relatively high (.930) that the lot will be accepted by chance. For higher values of p, the probability of lot acceptance by chance decreases. In addition, as p increases, the expected value moves away from the acceptable values, x≤ 2. This move reduces the chances of lot acceptance. 5.5 Solve the following problems by using the binomial formula. a. If n = 4 and p = .10, find P(x = 3). b. If n = 7 and p = .80, find P(x = 4). C. If n = 10 and p = .60, find P(x ≥ 7). d. If n = 12 and p = .45, find P(5 ≤ x ≤ 7). Solve the following problems by using the binomial tables (Table A.2). a. If n = 20 and p = .50, find P(x = 12). b. If n = 20 and p = .30, find P(x > 8). C. If n = 20 and p = .70, find P(x < 12). d. If n = 20 and p = 90, find 5.6
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