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\(X\) is a binomial random variable with the parameters shown. Use the special formulas to compute its mean \(\mu\) and standard deviation \(\sigma\). a. \(\quad n=14, p=0.55\) b. \(\quad n=83, p=0.05\) c. \(\quad n=957, p=0.35\) d. \(\quad n=1750, p=0.79\)

Short Answer

Expert verified
Mean and SD: a) 7.7, 1.86; b) 4.15, 1.99; c) 334.95, 14.74; d) 1382.5, 17.02.

Step by step solution

01

Understand the Binomial Distribution Parameters

In a binomial distribution, two main parameters define the random variable: \(n\), the number of trials, and \(p\), the probability of success on each trial. Using these parameters, we can calculate the mean and standard deviation of the distribution.
02

Apply the Formula for Mean

The mean of a binomial distribution is calculated using the formula \(\mu = n \times p\). Here, you will calculate the mean for each of the given scenarios using their respective \(n\) and \(p\) values:a. \(\mu = 14 \times 0.55 = 7.7\)b. \(\mu = 83 \times 0.05 = 4.15\)c. \(\mu = 957 \times 0.35 = 334.95\)d. \(\mu = 1750 \times 0.79 = 1382.5\)
03

Apply the Formula for Standard Deviation

The standard deviation of a binomial distribution is calculated using the formula \(\sigma = \sqrt{n \times p \times (1 - p)}\). Now, we find the standard deviation for each scenario:a. \(\sigma = \sqrt{14 \times 0.55 \times (1 - 0.55)} = \sqrt{14 \times 0.55 \times 0.45} = \sqrt{3.465} \approx 1.86\)b. \(\sigma = \sqrt{83 \times 0.05 \times (1 - 0.05)} = \sqrt{83 \times 0.05 \times 0.95} = \sqrt{3.9425} \approx 1.99\)c. \(\sigma = \sqrt{957 \times 0.35 \times (1 - 0.35)} = \sqrt{957 \times 0.35 \times 0.65} = \sqrt{217.2825} \approx 14.74\)d. \(\sigma = \sqrt{1750 \times 0.79 \times (1 - 0.79)} = \sqrt{1750 \times 0.79 \times 0.21} = \sqrt{290.085} \approx 17.02\)
04

Conclusion of Calculations

The mean and standard deviation for each scenario are:a. Mean = 7.7, Standard Deviation \(\approx 1.86\)b. Mean = 4.15, Standard Deviation \(\approx 1.99\)c. Mean = 334.95, Standard Deviation \(\approx 14.74\)d. Mean = 1382.5, Standard Deviation \(\approx 17.02\)

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Key Concepts

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

Mean and Standard Deviation
In probability theory, the concepts of mean and standard deviation are fundamental because they describe key features of a distribution. Let's define what these are in the context of a binomial distribution.

The **mean** of a distribution is essentially the average outcome you'd expect from a set of trials. For a binomial distribution, the formula for the mean is \[ \mu = n \times p \]where:
  • \( n \) is the number of trials
  • \( p \) is the probability of success on each trial
This formula calculates the central tendency of a binomial random variable; it tells us how many successes we expect to see.

The **standard deviation**, on the other hand, measures the spread or variability of the distribution. It answers the question of how much the number of successes varies from the mean. The standard deviation for a binomial distribution is calculated with the formula:\[ \sigma = \sqrt{n \times p \times (1 - p)} \]Here, the term \((1 - p)\) represents the probability of failure. This expression gives us insight into the fluctuation expected around the mean when the trials are conducted multiple times.
Binomial Random Variable
A **binomial random variable** is a type of discrete random variable that arises in binomial experiments. These are experiments that consist of a fixed number of independent trials, each with two possible outcomes: success or failure. Examples include flipping a coin, passing/failing an exam, or any scenario where there is a clear success/failure result.

The key characteristics of a binomial random variable are:
  • The number of trials, \( n \)
  • The probability of success on a single trial, \( p \)
These characteristics allow us to calculate the probability of observing any particular number of successes in our trials. With parameters \( n \) and \( p \), one can determine all probabilities concerning the number of successes, from zero to \( n \). Each possible outcome, given by the number of successes, constitutes a binomial random variable value.

The calculation of mean and standard deviation that we discussed earlier starts specifically from these two critical parameters, making it essential to fully understand "what" a binomial random variable describes and "how" these parameters relate to its determination.
Probability Theory
**Probability theory** is the branch of mathematics that deals with the analysis of random phenomena. In simple terms, it is the study of likelihood and uncertainty, providing a mathematical framework to quantify uncertain events or outcomes.

In the context of binomial distribution, probability theory helps us understand how often a particular number of successes might occur given a set number of trials and the probability of success in each individual trial. For example, if you wanted to know the likelihood of getting exactly three heads when flipping a fair coin five times, probability theory and the binomial distribution model come into play.

The core idea here is to determine the probability of each possible outcome in a random process. These probabilities form the basis for statistical inference, where predictions and deductions about broader datasets can be made by modeling them as random variables with known distributions. This theory is not only fundamental for binomial distributions but is also vital in all facets of statistics and forms the backbone of more advanced concepts.

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Most popular questions from this chapter

Classify each random variable as either discrete or continuous. a. The number of boys in a randomly selected three-child family. b. The temperature of a cup of coffee served at a restaurant. c. The number of no-shows for every 100 reservations made with a commercial airline. d. The number of vehicles owned by a randomly selected household. e. The average amount spent on electricity each July by a randomly selected household in a certain state.

A corporation has advertised heavily to try to insure that over half the adult population recognizes the brand name of its products. In a random sample of 20 adults, 14 recognized its brand name. What is the probability that 14 or more people in such a sample would recognize its brand name if the actual proportion \(p\) of all adults who recognize the brand name were only \(0.50 ?\)

An insurance company estimates that the probability that an individual in a particular risk group will survive one year is \(0.9825 .\) Such a person wishes to buy a \(\$ 150,000\) one-year term life insurance policy. Let \(C\) denote how much the insurance company charges such a person for such a policy. a. Construct the probability distribution of \(X\). (Two entries in the table will containC.) b. Compute the expected value \(E(x)\) of \(X\). c. Determine the value \(C\) must have in order for the company to break even on all such policies (that is, to average a net gain of zero per policy on such policies). d. Determine the value \(C\) must have in order for the company to average a net gain of \(\$ 250\) per policy on all such policies.

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A professional proofreader has a \(98 \%\) chance of detecting an error in a piece of written work (other than misspellings, double words, and similar errors that are machine detected). A work contains four errors. a. Find the probability that the proofreader will miss at least one of them. b. Show that two such proofreaders working independently have a \(99.96 \%\) chance of detecting an error in a piece of written work. c. Find the probability that two such proofreaders working independently will miss at least one error in a work that contains four errors.

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