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Chapter 6: Problem 5

The expected number of successes in a binomial experiment with \(n\) trials and probability of success \(p\) is ______.

Short Answer

Expert verified
The expected number of successes is given by \( n \times p \).

Step by step solution

01

- Understand the Binomial Distribution

A binomial distribution is used when there are exactly two mutually exclusive outcomes of a trial. The distribution tracks the number of successes in a given number of trials.
02

- Know the Formula for Expected Value

The formula for the expected number of successes in a binomial experiment is given by the product of the number of trials, n, and the probability of success, p.
03

- Apply the Formula

Multiply the number of trials by the probability of success to find the expected number of successes. This can be expressed as: \[ E(X) = n \times p \]

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Key Concepts

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

Binomial Distribution
The binomial distribution is a probability distribution that summarizes the likelihood that a value will take one of two independent states. In a binomial experiment, there are exactly two possible outcomes for each trial: a success or a failure.

Here are some essential points to remember:
  • Each trial is independent of the other trials.
  • The probability of success denoted by \( p \), remains constant in every trial.
For example, flipping a coin is a binomial experiment if you label 'heads' as a success and 'tails' as a failure.

When you perform a fixed number of trials \( n \), the binomial distribution helps you find the number of successes.
Expected Value Formula
The expected value (or mean) in the context of a binomial distribution refers to the average number of successes you can expect in a given number of trials.

The formula to calculate the expected value \( E(X) \) in a binomial experiment is straightforward. It is the product of the number of trials \( n \) and the probability of success \( p \). This can be written as:

\[ E(X) = n \times p \]

To help understand this better:
  • If you flip a coin 10 times \( (n = 10) \) with the probability of getting heads \( (p = 0.5) \), the expected value of heads is: \[ E(X) = 10 \times 0.5 = 5 \]
Probability of Success
In a binomial distribution, the probability of success is the probability that one trial will result in a success.

The probability of success is denoted by \( p \). It is essential that \( p \) stays the same from trial to trial in a binomial experiment.

Here are key points to take away:
  • The value of \( p \) is between 0 and 1.
  • The probability of failure is \( 1 - p \).
For instance, if you are rolling a fair six-sided die and you consider rolling a 4 as a success, then the probability of success \( p \) is \[ p = \frac{1}{6} \].

Understanding this concept helps you effectively apply other related formulas like the expected value formula.

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Most popular questions from this chapter

(a) construct a binomial probability distribution with the given parameters; (b) compute the mean and standard deviation of the random variable using the methods of Section \(6.1 ;\) (c) compute the mean and standard deviation, using the methods of this section; and \((d)\) draw a graph of the probability distribution and comment on its shape. $$ n=10, p=0.2 $$

A binomial probability experiment is conducted with the given parameters. Compute the probability of \(x\) successes in the \(n\) independent trials of the experiment. $$ n=50, p=0.02, x=3 $$

According to www.meretrix.com, airline fatalities occur at the rate of 0.05 fatal accidents per 100 million miles. Find the probability that, during the next 100 million miles of flight, there will be (a) exactly zero fatal accidents. Interpret the result. (b) at least one fatal accident. Interpret the result. (c) more than one fatal accident. Interpret the result.

Data from the National Center for Health Statistics show that spina bifida occurs at the rate of 28 per 100,000 live births. Let the random variable \(X\) represent the number of occurrences of spina bifida in a random sample of 100,000 live births. (a) What is the expected number of children with spina bifida per 100,000 live births in any given year? (b) Using statistical software such as Minitab, simulate taking 200 random samples of 100,000 live births, assuming \(\mu=28\) (c) Approximate the probability that fewer than 18 births per 100,000 result in spina bifida. (d) In 2005,17.96 births per 100,000 resulted in babies born with spina bifida. In light of the results of parts (b) and (c), is this an unusual occurrence? What might you conclude?

A life insurance company sells a \(\$ 250,000\) 1-year term life insurance policy to a 20-year-old female for \$200. According to the National Vital Statistics Report, 58(21) , the probability that the female survives the year is 0.999544 . Compute and interpret the expected value of this policy to the insurance company.
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