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Let \(x\) be a binomial random variable with \(n=7\) and \(p=.5 .\) Find the values of the quantities in Exercises \(11-15 .\) $$ \sigma=\sqrt{n p q} $$

Short Answer

Expert verified
Answer: The standard deviation of the given binomial random variable is approximately \(1.32\).

Step by step solution

01

Find the value of q

Since \(p=0.5\), \(q = 1-p = 1-0.5 = 0.5\).
02

Apply the standard deviation formula

Now, we plug \(n=7\), \(p=0.5\), and \(q=0.5\) into the formula \(\sigma = \sqrt{npq}\). This gives us: $$\sigma = \sqrt{(7)(0.5)(0.5)}$$.
03

Calculate the standard deviation

Perform the multiplications and square root calculation: $$\sigma = \sqrt{(7)(0.25)} = \sqrt{1.75} \approx 1.32$$. So, the standard deviation of the given binomial random variable is approximately \(1.32\).

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Key Concepts

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

Binomial Random Variable
A binomial random variable is a specific type of variable used in statistics that models the number of successes in a fixed number of independent trials, with each trial having the same probability of success.

For instance, if you flip a coin 7 times, each flip represents an independent trial with two possible outcomes, success (heads) or failure (tails). If we define success as getting 'heads', and we're interested in the number of 'heads' we might get out of the 7 flips, this count is our binomial random variable.

In a binomial setting, the probability of success is denoted by the letter 'p', while 'q' represents the probability of failure (where q = 1 - p). The number of trials is represented by 'n'. Therefore, if we designate x as a binomial random variable with n trials and a probability p of success on each trial, we say x follows a binomial distribution with parameters n and p, expressed as B(n, p).
Probability Distribution
A probability distribution is a mathematical function that gives the probabilities of occurrence of different possible outcomes for an experiment. It's a statistical tool that helps us understand the likelihood of various results.

For a binomial random variable, the probability distribution is the binomial distribution. It tells us how likely it is to achieve a certain number of successes in a fixed number of trials. This distribution is governed by the parameters n (number of trials) and p (probability of success).

To use the binomial distribution, we calculate the probability of getting exactly k successes (k is any number from 0 to n) using the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
Here, \( \binom{n}{k} \) is a binomial coefficient, which calculates the number of ways k successes can occur in n trials, \( p^k \) is the probability of having k successes, and \( q^{n-k} \) is the probability of the remaining trials resulting in failure.
Standard Deviation Formula
The standard deviation is a measure of how spread out numbers are in a data set. In the context of a binomial distribution, it quantifies the variability of the number of successes one can expect.

To calculate the standard deviation of a binomial random variable, we use the formula: \[ \sigma = \sqrt{n p q} \
where \( \sigma \) represents the standard deviation, \( n \) is the number of trials, \( p \) is the probability of success, and \( q \) is the probability of failure (q = 1 - p).

For the exercise given, where \( n = 7 \) and \( p = 0.5 \), the calculation of the standard deviation involves first determining q as \( 1 - p \) and then substituting the values into the formula. After solving the arithmetic inside the square root, we find that \( \sigma \) is approximately 1.32, which tells us about the average variability in the number of 'heads' we might see when we flip a coin 7 times.

Understanding this formula and its components is crucial for interpreting the variability and consistency of results across multiple trials in binomial experiments.

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Let \(x\) be the number of successes observed in a sample of \(n=5\) items selected from a population of \(N=10 .\) Suppose that of the \(N=10\) items, \(M=6\) are considered "successes." Find the probabilities in Exercises \(11-13 .\) The probability of observing at least two successes.

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