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Chapter 5: Problem 1

Consider a binomial random variable with \(n=8\) and \(p=.7\). Let \(x\) be the number of successes in the sample. a. Find the probability that \(x\) is 3 or less. b. Find the probability that \(x\) is 3 or more. c. Find \(P(x<3)\). d. Find \(P(x=3)\). e. Find \(P(3 \leq x \leq 5)\).

Short Answer

Expert verified
Answer: \(P(x \leq 3) \approx 0.0612\) 2. What is the probability that the number of successes (x) is 3 or more? Answer: \(P(x \geq 3) \approx 0.9388\) 3. What is the probability that the number of successes (x) is less than 3? Answer: \(P(x < 3) \approx 0.0183\) 4. What is the probability that the number of successes (x) is equal to 3? Answer: \(P(x = 3) \approx 0.0429\) 5. What is the probability that the number of successes (x) is between 3 and 5, inclusive? Answer: \(P(3 \leq x \leq 5) \approx 0.2785\)

Step by step solution

01

Find the probability that \(x\) is 3 or less.

To find the probability that \(x\) is 3 or less, we need to calculate the sum of the probabilities for \(x=0\), \(x=1\), \(x=2\) and \(x=3\). We can do this using the binomial probability formula: \(P(x \leq 3) = P(x=0) + P(x=1) + P(x=2) + P(x=3)\) For each value of \(x\), we can plug in the values for \(n\), \(p\), and \(x\) into the formula: \(P(x \leq 3) = \binom{8}{0} (0.7)^0 (0.3)^8 + \binom{8}{1} (0.7)^1 (0.3)^7 + \binom{8}{2} (0.7)^2 (0.3)^6 + \binom{8}{3} (0.7)^3 (0.3)^5\) Calculate the probabilities and sum them up: \(P(x \leq 3) \approx 0.00034 + 0.0033 + 0.0147 + 0.0429 \approx 0.0612\)
02

Find the probability that \(x\) is 3 or more.

Since the sum of all probabilities in a binomial distribution is 1, we can find the probability that \(x\) is 3 or more by subtracting the probability that \(x\) is less than 3 from 1: \(P(x \geq 3) = 1 - P(x < 3)\) We already found that \(P(x \leq 3) \approx 0.0612\), so: \(P(x \geq 3) \approx 1 - 0.0612 \approx 0.9388\)
03

Find \(P(x < 3)\).

To find the probability that \(x\) is less than 3, we need to calculate the sum of the probabilities for \(x=0\), \(x=1\), and \(x=2\). We can do this using the binomial probability formula: \(P(x < 3) = P(x=0) + P(x=1) + P(x=2)\) For each value of \(x\), we can plug in the values for \(n\), \(p\), and \(x\) into the formula: \(P(x < 3) = \binom{8}{0} (0.7)^0 (0.3)^8 + \binom{8}{1} (0.7)^1 (0.3)^7 + \binom{8}{2} (0.7)^2 (0.3)^6\) Calculate the probabilities and sum them up: \(P(x < 3) \approx 0.00034 + 0.0033 + 0.0147 \approx 0.0183\)
04

Find \(P(x = 3)\).

We already found the probability that \(x = 3\) in Step 1. It is approximately: \(P(x = 3) \approx 0.0429\)
05

Find \(P(3 \leq x \leq 5)\).

To find the probability that \(x\) is between 3 and 5 inclusive, we need to calculate the sum of the probabilities for \(x=3\), \(x=4\), and \(x=5\). We can do this using the binomial probability formula: \(P(3 \leq x \leq 5) = P(x=3) + P(x=4) + P(x=5)\) For each value of \(x\), we can plug in the values for \(n\), \(p\), and \(x\) into the formula: \(P(3 \leq x \leq 5) = \binom{8}{3} (0.7)^3 (0.3)^5 + \binom{8}{4} (0.7)^4 (0.3)^4 + \binom{8}{5} (0.7)^5 (0.3)^3\) Calculate the probabilities and sum them up: \(P(3 \leq x \leq 5) \approx 0.0429 + 0.0895 + 0.1461 \approx 0.2785\)

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