5. Suppose 100 steel rods are chosen at random from a large box of steel rods (more than 10,000). Suppose it is estimated that 5% are defective. The probabilities of the following events using the binomial distribution are below • P(A: none of the 100 widgets is defective) = 0.0059 • P(B: exactly 1 is defective) = 0.0371 • P(C: no more than 2 are defective) = 0.1183 • P(D: exactly 5 are defective) = 0.1800 • P(E: more than 5 are defective) <2.1*10-8 Use the Poison distribution to compute approximations to the same probabilities using a lambda value of 5 (.05*100)
5. Suppose 100 steel rods are chosen at random from a large box of steel rods (more than 10,000). Suppose it is estimated that 5% are defective. The probabilities of the following events using the binomial distribution are below • P(A: none of the 100 widgets is defective) = 0.0059 • P(B: exactly 1 is defective) = 0.0371 • P(C: no more than 2 are defective) = 0.1183 • P(D: exactly 5 are defective) = 0.1800 • P(E: more than 5 are defective) <2.1*10-8 Use the Poison distribution to compute approximations to the same probabilities using a lambda value of 5 (.05*100)
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Chapter1: Starting With Matlab
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
Transcribed Image Text:**Problem 5: Probability Analysis Using Binomial and Poisson Distributions**
*Scenario:*
Assume 100 steel rods are selected randomly from a large box containing more than 10,000 steel rods. It is estimated that 5% of these rods are defective. The probabilities for various events are calculated using the binomial distribution.
*Probabilities:*
- **P(A):** Probability that none of the 100 rods is defective = 0.0059
- **P(B):** Probability that exactly 1 rod is defective = 0.0371
- **P(C):** Probability that no more than 2 rods are defective = 0.1183
- **P(D):** Probability that exactly 5 rods are defective = 0.1800
- **P(E):** Probability that more than 5 rods are defective < 2.1 * 10⁻⁸
*Task:*
Use the Poisson distribution to compute approximations for the same probabilities. Here, use a lambda value (λ) of 5, calculated as 5% of 100 (0.05 * 100).

Transcribed Image Text:e. P(E: more than 5 are defective)
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