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Consider a binomial experiment with \(n=10\) and \(p=.10\) a. Compute \(f(0)\) b. Compute \(f(2)\) c. Compute \(P(x \leq 2)\) d. Compute \(P(x \geq 1)\) e. Compute \(E(x)\) f. Compute \(\operatorname{Var}(x)\) and \(\sigma\).

Short Answer

Expert verified
a. 0.3487, b. 0.1938, c. 0.9299, d. 0.6513, e. 1, f. Var = 0.9, σ ≈ 0.9487.

Step by step solution

01

Compute f(0)

To find the probability of zero successes, denoted as \( f(0) \), use the binomial probability formula: \[ f(x) = \binom{n}{x} p^x (1-p)^{n-x} \]where \( n = 10 \), \( p = 0.10 \), and \( x = 0 \). So, \[ f(0) = \binom{10}{0} (0.10)^0 (0.90)^{10} = 1 \times 1 \times 0.3487 = 0.3487 \]
02

Compute f(2)

We use the same binomial formula to find \( f(2) \). \[ f(2) = \binom{10}{2} (0.10)^2 (0.90)^8 \]First, calculate the binomial coefficient:\[ \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \]Now plug it into the formula:\[ f(2) = 45 \times 0.01 \times 0.4305 = 0.1938 \]
03

Compute P(x ≤ 2)

To find the cumulative probability of \( x \) being less than or equal to 2, sum \( f(0), f(1), \text{ and } f(2) \).We already have \( f(0) = 0.3487 \) and \( f(2) = 0.1938 \).Now compute \( f(1) \):\[ f(1) = \binom{10}{1} (0.10)^1 (0.90)^9 \]\[ \binom{10}{1} = 10 \]\[ f(1) = 10 \times 0.10 \times 0.3874 = 0.3874 \]Thus, \[ P(x \leq 2) = f(0) + f(1) + f(2) = 0.3487 + 0.3874 + 0.1938 = 0.9299 \]
04

Compute P(x ≥ 1)

To determine \( P(x \geq 1) \), recognize that this is the complement of \( P(x = 0) \). Thus, \[ P(x \geq 1) = 1 - P(x = 0) = 1 - 0.3487 = 0.6513 \]
05

Compute E(x)

The expected value \( E(x) \) for a binomial distribution is given by \[ E(x) = n \times p \]Substituting the values of \( n \) and \( p \):\[ E(x) = 10 \times 0.10 = 1 \]
06

Compute Var(x) and σ

For a binomial distribution, variance \( \operatorname{Var}(x) \) is calculated as \[ \operatorname{Var}(x) = n \times p \times (1-p) \]Substitute the values:\[ \operatorname{Var}(x) = 10 \times 0.10 \times 0.90 = 0.9 \]The standard deviation \( \sigma \) is the square root of the variance:\[ \sigma = \sqrt{0.9} \approx 0.9487 \]

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Key Concepts

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

Binomial Probability Formula
In a binomial experiment, we often want to know the probability of getting a certain number of successes in a series of independent experiments, like flipping a coin multiple times. The binomial probability formula helps us find this probability. The formula is:\[ f(x) = \binom{n}{x} p^x (1-p)^{n-x} \]- *\( n \)* is the number of experiments or trials.- *\( x \)* is the exact number of successes we're interested in.- *\( p \)* is the probability of success on a single trial.The term \( \binom{n}{x} \) is known as the binomial coefficient, representing the number of ways to choose \( x \) successes from \( n \) trials. Probabilities \( p^x \) and \( (1-p)^{n-x} \) account for the likelihood of these successes and failures, respectively.
By multiplying these values, we calculate the probability of observing exactly \( x \) successes in \( n \) trials.
Expected Value
The expected value in probability theory gives us a measure of the central tendency or average outcome of a random experiment. For a binomial distribution, the expected value \( E(x) \) helps answer the question: *"On average, how many successes can we expect?"*
The formula to calculate the expected value for a binomial distribution is:\[ E(x) = n \times p \]- Here, *\( n \)* represents the total number of trials, and *\( p \)* is the probability of a single success.By simply multiplying these two values, we arrive at the expected number of successes over all trials.
This calculation is useful in anticipating results in practical scenarios. For instance, if you expect to roll a dice 60 times hoping to land on a specific number, you can use this formula to predict how many times the desired number might appear.
Variance
Variance is a key concept in statistics, measuring the dispersion or spread of random variables around the expected value. In a binomial distribution, variance provides insight into how much the results might vary from the average or mean.
The formula for computing the variance \( \operatorname{Var}(x) \) in a binomial distribution is:\[ \operatorname{Var}(x) = n \times p \times (1-p) \]Here's what each component means:- *\( n \)* is the number of trials.- *\( p \)* is the probability of success.- *\((1-p)\)* is the probability of failure.The variance can help assess the reliability or consistency of the outcomes over several trials.
A higher variance indicates greater variability, suggesting that the actual number of successes could significantly differ from the expected value.
Standard Deviation
Standard deviation is like the "average distance" each result is from the expected value. It helps make sense of variance by translating spread information from a squared to a linear scale. For binomial distributions, it is the square root of the variance.
The formula to find the standard deviation \( \sigma \) is:\[ \sigma = \sqrt{\operatorname{Var}(x)} \]Let's break this down:- First, calculate the variance \( \operatorname{Var}(x) \) using the binomial variance formula.- Then, take the square root of that variance to obtain the standard deviation.Standard deviation adds practical insights: It shows what typical variations from the average look like in regular terms.
A smaller standard deviation suggests tighter clustering of results around the expected value, while a larger one implies more considerable variation.

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Most popular questions from this chapter

Customer arrivals at a bank are random and independent; the probability of an arrival in any one-minute period is the same as the probability of an arrival in any other one-minute period. Answer the following questions, assuming a mean arrival rate of three customers per minute. a. What is the probability of exactly three arrivals in a one-minute period? b. What is the probability of at least three arrivals in a one-minute period?

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