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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
Based on the given information (n=7 trials, p=0.3 probability of success), calculate: a. The probability of having exactly 4 successes: P(X=4) ≈ 0.0975 b. The probability of having at most 1 success: P(X ≤ 1) ≈ 0.3525 c. The probability of having more than 1 success: P(X > 1) ≈ 0.6475 d. The mean of the binomial distribution: μ ≈ 2.1 e. The standard deviation of the binomial distribution: σ ≈ 1.229

Step by step solution

01

a. P(x=4) - Calculate the probability of having exactly 4 successes

To find the probability of having exactly 4 successes in 7 trials, we will use the binomial probability formula: \(P(X=k) = {n\choose k} p^k q^{n-k}\) where \(n=7\), \(k=4\), \(p = 0.3\) and \(q = 0.7\). Plug these values into the formula: \(P(X=4) = {7\choose 4} (0.3)^4 (0.7)^3\) Now, calculate the binomial coefficient: \({7\choose 4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = 35\) Finally, plug in the values and calculate the probability: \(P(X=4) = 35 (0.3)^4 (0.7)^3 \approx 0.0975\)
02

b. P(x ≤ 1) - Calculate the probability of having at most 1 success

To find the probability of having at most 1 success in 7 trials, we need to find the sum of probabilities for exactly 0 successes and exactly 1 success: \(P(X \leq 1) = P(X=0) + P(X=1)\) Using the binomial probability formula, calculate each probability: \(P(X=0) = {7\choose 0} (0.3)^0 (0.7)^7\) \(P(X=1) = {7\choose 1} (0.3)^1 (0.7)^6\) Compute the binomial coefficients: \({7\choose 0} = 1\) \({7\choose 1} = 7\) Now, plug in the values and calculate the probabilities: \(P(X=0) = 1 (0.3)^0 (0.7)^7 \approx 0.0824\) \(P(X=1) = 7 (0.3)^1 (0.7)^6 \approx 0.2701\) Finally, sum the probabilities: \(P(X \leq 1) = 0.0824 + 0.2701 \approx 0.3525\)
03

c. P(x > 1) - Calculate the probability of having more than 1 success

Considering that the total probability equals 1, we can calculate the probability of having more than 1 success as the complementary probability: \(P(X > 1) = 1 - P(X \leq 1)\) Use the previously calculated result for \(P(X \leq 1)\): \(P(X > 1) = 1 - 0.3525 \approx 0.6475\)
04

d. μ = np - Calculate the mean of the binomial distribution

To calculate the mean of the binomial distribution, use the formula \(\mu = np\): \(\mu = 7 \times 0.3 = 2.1\)
05

e. σ = √(npq) - Calculate the standard deviation of the binomial distribution

To calculate the standard deviation of the binomial distribution, use the formula \(\sigma = \sqrt{npq}\): \(\sigma = \sqrt{7 \times 0.3 \times 0.7} \approx 1.229\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Calculation
Understanding how to calculate probabilities in a binomial distribution is crucial. In a binomial setting, we are interested in the outcomes of a fixed number of trials, each with two possible results: success or failure. The binomial probability formula helps determine the probability of achieving exactly a certain number of successes in a given number of trials, with success probability denoted as \(p\), and failure probability as \(q = 1 - p\). For example, calculating \(P(X = 4)\) involves calculating the chance of getting exactly four successes out of seven trials, using:
  • The number of trials \(n\) (7 in this example)
  • The number of successes \(k\) (4 here)
  • Probability of success \(p\) (0.3 here)
  • Using the binomial coefficient and probability formula: \(P(X=k) = {n\choose k} p^k q^{n-k}\).
By plugging these values into the equation, you can find the exact probabilities for varying numbers of successes.
Binomial Coefficient
The binomial coefficient \({n\choose k}\) is a key component in calculating binomial probabilities. It represents the number of ways to choose \(k\) successes from \(n\) trials. Mathematically, it is expressed as:\[{n\choose k} = \frac{n!}{k!(n-k)!}\]In simpler terms, it is the number of combinations possible, not considering the order of the outcomes. For example, to calculate \({7\choose 4}\), which is needed for finding \(P(X=4)\), you compute using factorials:
  • Factorial of 7: \(7!\)
  • Factorial of 4: \(4!\)
  • Factorial of 3: \(3!\)
Applying the formula, you get the number of combinations as 35. Binomial coefficients are essential in determining probabilities in a binomial distribution.
Mean of a Binomial Distribution
The mean (or expected value) of a binomial distribution provides the average outcome you can anticipate from a series of trials. It is computed using the formula:\[\mu = np\]where \(n\) is the number of trials, and \(p\) is the probability of success in each trial. This calculation gives us the balance point of the distribution. For instance, with \(n=7\) and \(p=0.3\), the mean \(\mu\) calculates to:
  • \(\mu = 7 \times 0.3 = 2.1\)
This tells you that, on average, you can expect about 2.1 successes in 7 trials. Understanding the mean helps in setting expectations for the outcome of binomial experiments.
Standard Deviation of a Binomial Distribution
The standard deviation of a binomial distribution shows the variability or spread of the distribution around the mean. It indicates how much individual results can deviate from the average. The standard deviation is calculated using the formula:\[\sigma = \sqrt{npq}\]where:
  • \(n\) is the number of trials
  • \(p\) is the probability of success
  • \(q = 1 - p\) is the probability of failure
For \(n = 7\), \(p = 0.3\), and \(q = 0.7\), the standard deviation \(\sigma\) is:
  • \(\sigma = \sqrt{7 \times 0.3 \times 0.7} \approx 1.229\)
This means that the actual number of successes will typically vary from the mean by about 1.229 successes. Understanding standard deviation is crucial for assessing the consistency of results in repeated trials.

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Most popular questions from this chapter

One model for plant competition assumes that there is a zone of resource depletion around each plant seedling. Depending on the size of the zones and the density of the plants, the zones of resource depletion may overlap with those of other seedlings in the vicinity. When the seeds are randomly dispersed over a wide area, the number of neighbors that a seedling may have usually follows a Poisson distribution with a mean equal to the density of seedlings per unit area. Suppose that the density of seedlings is four per square meter \(\left(\mathrm{m}^{2}\right)\). a. What is the probability that a given seedling has no neighbors within \(1 \mathrm{~m}^{2} ?\) b. What is the probability that a seedling has at most three neighbors per \(\mathrm{m}^{2}\) ? c. What is the probability that a seedling has five or more neighbors per \(\mathrm{m}^{2} ?\) d. Use the fact that the mean and variance of a Poisson random variable are equal to find the proportion of neighbors that would fall into the interval \(\mu \pm 2 \sigma .\) Comment on this result.

Under what conditions can the Poisson random variable be used to approximate the probabilities associated with the binomial random variable? What application does the Poisson distribution have other than to estimate certain binomial probabilities?

Let \(x\) be a binomial random variable with \(n=10\) and \(p=.4 .\) Find these values: a. \(P(x=4)\) b. \(P(x \geq 4)\) c. \(P(x>4)\) d. \(P(x \leq 4)\) e. \(\mu=n p\) f. \(\sigma=\sqrt{n p q}\)

Use the formula for the binomial probability distribution to calculate the values of \(p(x)\) and construct the probability histogram for \(x\) when \(n=6\) and \(p=.2\). [HINT: Calculate \(P(x=k)\) for seven different values of \(k .]\)

Consider a Poisson random variable with \(\mu=2.5 .\) Use the Poisson formula to calculate the following probabilities: a. \(P(x=0)\) b. \(P(x=1)\) c. \(P(x=2)\) d. \(P(x \leq 2)\)

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