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In Exercises \(\mathrm{P} .112\) to \(\mathrm{P} .115,\) calculate the mean and standard deviation of the binomial random variable. A binomial random variable with } n=800 \text { and }\\\ &p=0.25 \end{aligned} $$

Short Answer

Expert verified
The mean of the binomial distribution is 200 and the standard deviation is approximately 12.25.

Step by step solution

01

Calculate the mean

The mean of a binomial distribution is calculated using the following formula: \\[µ = np\\]. Here, n is the number of trials and p is the probability of success. In this case, given numbers are n=800 and p=0.25, so the mean would be: \\[µ = n * p = 800 * 0.25 = 200\\]
02

Calculate the standard deviation

The standard deviation of a binomial distribution is calculated using the following formula: \\[σ = sqrt {np(1 - p)}\\]. Here, n and p are defined as before while (1 - p) denotes the probability of failure. Plugging in the given values, the standard deviation would be: \\[σ = sqrt {800 * 0.25 * (1 - 0.25)} = sqrt {150} ≈ 12.25\\].

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Key Concepts

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

Mean of Binomial Distribution
Understanding the mean of a binomial distribution is crucial when dealing with probabilities in experiments or processes that have two possible outcomes, like success or failure. The mean, often represented by the Greek letter µ (mu), is calculated by multiplying the total number of trials () by the probability of success ((++)).

In simpler terms, it tells us the expected number of successes in a given number of trials. For instance, if we flip a coin 100 times (n=100) and we're interested in the number of times it lands heads (assuming the probability of heads is 0.5), the mean number of heads would be 50 (mean = 100 * 0.5).

Applying this to a classroom scenario, if a student answers 800 multiple-choice questions (n=800) and the probability of choosing the correct answer by guessing is 0.25 (p=0.25), the mean number of correct answers would be 200. It becomes a useful figure for predicting outcomes over the long run.
Standard Deviation of Binomial Distribution
The standard deviation in a binomial distribution, represented as (σ), measures how spread out the numbers are in your set of data. It tells us roughly how far the outcomes will vary from the mean. A higher standard deviation indicates that the results are more spread out.

The formula for calculating the standard deviation is (σ = sqrt {np(1 - p)}). It takes into account the number of trials (n), the probability of success (p), and the probability of failure (1 - p). So, using our previous classroom example with n=800 trials and p=0.25, the standard deviation is about 12.25. This value gives us an insight into the variability or consistency of the student's success rate when answering multiple-choice questions by guessing.
Binomial Random Variable
A binomial random variable is the cornerstone of binomial distribution problems. It counts the number of successes in a series of trials where each trial has only two possible outcomes - often called 'success' and 'failure'. For a variable to be binomial, trials must be independent, and the probability of success must remain the same across each trial.

As an example, if a basketball player takes 20 free throws, and the probability of making a free throw is constant, say 0.6, the number of free throws the player makes is a binomial random variable. This is because each throw is independent (the outcome of one doesn't affect the other), and the probability remains the same (0.6) for each free throw.
Probability of Success
In the context of binomial distributions, the probability of success (denoted as p) is the chance of a single trial resulting in a success. It's a value between 0 and 1 that stays constant throughout all trials. The probability of failure is simply (1 - p).

For instance, if the probability of a new business succeeding within its first year is 0.10, then p=0.10. When this probability is used in computation of a binomial distribution, it allows for the assessment of outcomes, such as forecasting how many businesses out of a hundred might succeed. Understanding the probability of success is essential for accurate predictions of the binomial distribution metrics.

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Most popular questions from this chapter

Average Household Size for Renter-Occupied Units Table \(\mathrm{P} .12\) in Exercise \(\mathrm{P} .83\) gives the probability function for the random variable giving the household size for a renter-occupied housing unit in the US. (a) Find the mean household size. (b) Find the standard deviation for household size.

Use the fact that we have independent events \(\mathrm{A}\) and \(\mathrm{B}\) with \(P(A)=0.7\) and \(P(B)=0.6\). Find \(P(A\) or \(B)\).

In a bag of peanut \(M\) \& M's, there are \(80 \mathrm{M} \& \mathrm{Ms}\), with 11 red ones, 12 orange ones, 20 blue ones, 11 green ones, 18 yellow ones, and 8 brown ones. They are mixed up so that each candy piece is equally likely to be selected if we pick one. (a) If we select one at random, what is the probability that it is red? (b) If we select one at random, what is the probability that it is not blue? (c) If we select one at random, what is the probability that it is red or orange? (d) If we select one at random, then put it back, mix them up well (so the selections are independent) and select another one, what is the probability that both the first and second ones are blue? (e) If we select one, keep it, and then select a second one, what is the probability that the first one is red and the second one is green?

Find endpoint(s) on the given normal density curve with the given property. P.147 (a) The area to the left of the endpoint on a \(N(100,15)\) curve is about 0.75 (b) The area to the right of the endpoint on a \(N(8,1)\) curve is about 0.03 .

Ask you to convert an area from one normal distribution to an equivalent area for a different normal distribution. Draw sketches of both normal distributions, find and label the endpoints, and shade the regions on both curves. The area above 2.1 for a standard normal distribution converted to a \(N(500,80)\) distribution

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