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Chapter 5: Problem 7

Let \(x\) be a binomial random variable with \(n=7\), \(p=.3 .\) Find these values: a. \(P(x=4)\) b. \(P(x \leq 1)\) c. \(P(x>1)\) d. \(\mu=n p\) e. \(\sigma=\sqrt{n p q}\)

Short Answer

Expert verified
Also, find the mean and standard deviation of the binomial random variable. Answer: The probabilities are as follows: a. The probability of having exactly 4 successes (P(x=4)) is approximately 0.130. b. The probability of having less than or equal to 1 success (P(x鈮�1)) is approximately 0.319. c. The probability of having greater than 1 success (P(x>1)) is approximately 0.681. d. The mean (渭) of this binomial random variable is 2.1. e. The standard deviation (蟽) of this binomial random variable is approximately 1.2.

Step by step solution

01

Understanding the binomial probability formula

The binomial probability formula is used to calculate the probability of a particular outcome for a binomial random variable. It is given by \(P(x=k) = \binom{n}{k}p^k(1-p)^{n-k}\), where \(n\) is the number of trials, \(k\) is the number of desired successes, and \(p\) is the probability of success on a single trial. We will use this formula to compute the probability required in parts a, b and c.
02

Calculating \(P(x=4)\)

To calculate \(P(x=4)\), we use the binomial probability formula, where \(n=7, k=4,\) and \(p=0.3\). Therefore, \(P(x=4) = \binom{7}{4}(0.3)^4(1-0.3)^{7-4}\). Using the formula for combinations, we have: \(P(x=4) = \frac{7!}{4!(7-4)!}(0.3)^4(0.7)^3\) Computing the values, we get \(P(x=4) \approx 0.130\).
03

Calculating \(P(x \leq 1)\)

To calculate \(P(x \leq 1)\), we need to find the probability of \(x=0\) and \(x=1\) and then sum these probabilities. Using the binomial probability formula, we calculate: \(P(x=0) = \binom{7}{0}(0.3)^0(1-0.3)^{7-0}\) \(P(x=1) = \binom{7}{1}(0.3)^1(1-0.3)^{7-1}\) Adding these probabilities, we have: \(P(x \leq 1) = P(x=0) + P(x=1) \approx 0.082 + 0.237 \approx 0.319\).
04

Calculating \(P(x > 1)\)

To calculate \(P(x > 1)\), we need to find the probability of all the other outcomes and subtract it from 1. As we know \(P(x \leq 1)\), we simply calculate: \(P(x > 1) = 1 - P(x \leq 1) = 1 - 0.319 \approx 0.681\).
05

Calculate the mean \(\mu = np\)

We can find the mean of the binomial variable by using the formula \(\mu = np\). Here, \(n=7\) and \(p=0.3\), so the mean is: \(\mu = 7 \times 0.3 = 2.1\).
06

Calculate the standard deviation \(\sigma = \sqrt{np(1-p)}\)

We can find the standard deviation of the binomial variable by using the formula \(\sigma = \sqrt{np(1-p)}\). Here, we have \(n=7, p=0.3\), and \(1-p = 0.7\), so the standard deviation is: \(\sigma = \sqrt{7 \times 0.3 \times 0.7} \approx 1.2\). Now, we have the answers for all the parts: a. \(P(x=4) \approx 0.130\) b. \(P(x \leq 1) \approx 0.319\) c. \(P(x > 1) \approx 0.681\) d. \(\mu = 2.1\) e. \(\sigma \approx 1.2\)

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Most popular questions from this chapter

Suppose that \(10 \%\) of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in this area are randomly selected and checked for whitefly. a. What is the average number of fields sampled that are infested with whitefly? b. Within what limits would you expect to find the number of infested fields, with probability approximately \(95 \% ?\) c. What might you conclude if you found that \(x=25\) fields were infested? Is it possible that one of the characteristics of a binomial experiment is not satisfied in this experiment? Explain.

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