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Find the mean and standard deviation for a binomial distribution with \(n=100\) and these values of \(p\) : a. \(p=.01\) b. \(p=.9\) c. \(p=.3\) d. \(p=.7\) e. \(p=.5\)

Short Answer

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Question: For a binomial distribution with n=100, find the mean and standard deviation for each of the following values of p: a) 0.01 b) 0.9 c) 0.3 d) 0.7 e) 0.5. Answer: a) Mean = 1, Standard Deviation = √0.99 b) Mean = 90, Standard Deviation = √9 c) Mean = 30, Standard Deviation = √21 d) Mean = 70, Standard Deviation = √21 e) Mean = 50, Standard Deviation = √25

Step by step solution

01

a. Mean and Standard Deviation for p = 0.01

To calculate the mean and standard deviation for a binomial distribution with \(n=100\) and \(p=0.01\): Mean = \(100 \times 0.01 = 1\) Standard Deviation = \(\sqrt{100 \times 0.01 \times (1-0.01)} = \sqrt{1 \times 0.99} = \sqrt{0.99}\)
02

b. Mean and Standard Deviation for p = 0.9

To calculate the mean and standard deviation for a binomial distribution with \(n=100\) and \(p=0.9\): Mean = \(100 \times 0.9 = 90\) Standard Deviation = \(\sqrt{100 \times 0.9 \times (1-0.9)} = \sqrt{90 \times 0.1} = \sqrt{9}\)
03

c. Mean and Standard Deviation for p = 0.3

To calculate the mean and standard deviation for a binomial distribution with \(n=100\) and \(p=0.3\): Mean = \(100 \times 0.3 = 30\) Standard Deviation = \(\sqrt{100 \times 0.3 \times (1-0.3)} = \sqrt{30 \times 0.7} = \sqrt{21}\)
04

d. Mean and Standard Deviation for p = 0.7

To calculate the mean and standard deviation for a binomial distribution with \(n=100\) and \(p=0.7\): Mean = \(100 \times 0.7 = 70\) Standard Deviation = \(\sqrt{100 \times 0.7 \times (1-0.7)} = \sqrt{70 \times 0.3} = \sqrt{21}\)
05

e. Mean and Standard Deviation for p = 0.5

To calculate the mean and standard deviation for a binomial distribution with \(n=100\) and \(p=0.5\): Mean = \(100 \times 0.5 = 50\) Standard Deviation = \(\sqrt{100 \times 0.5 \times (1-0.5)} = \sqrt{50 \times 0.5} = \sqrt{25}\)

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Key Concepts

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

Mean of a Binomial Distribution
Understanding the mean of a binomial distribution is essential for grasping how probability and statistics interconnect. In a binomial distribution, the mean represents the average expected outcome and is determined by multiplying the number of trials (n) by the probability of success (p). The formula is expressed as \( \text{Mean} = n \times p \).

For instance, with a probability (p) of 0.3 of success on a single trial, and 100 trials (n), the mean would be 30, indicating that, on average, you would expect 30 successful outcomes. This value gives you a quick glimpse of what to expect from the random process described by the binomial distribution.
Standard Deviation of a Binomial Distribution
The standard deviation in a binomial distribution quantifies the variability or spread of the distribution around the mean. It measures how much, on average, each trial's result deviates from the expected mean. The mathematical formula for the standard deviation (\( \sigma \)) of a binomial distribution is \( \sigma = \sqrt{n \times p \times (1-p)} \).

Using this formula, if we have a binomial distribution with a probability of success of 0.7 per trial and 100 trials, the standard deviation would be \( \sigma = \sqrt{100 \times 0.7 \times 0.3} \), which calculates to the square root of 21. This tells us how widely the number of successes might vary from the expected mean.
Probability and Statistics
Probability and statistics are two fundamental aspects of understanding binomial distributions. Probability is used to predict the likelihood of an event occurring, while statistics is the practice of collecting, analyzing, and interpreting data to make those predictions. The binomial distribution is a probability distribution that summarizes the likelihood that a value will take on one of two independent values under a given set of parameters or conditions.

This concept is crucial in statistical applications like quality control, survey analysis, and many other fields where the outcomes are binary and probabilities are constant across trials. Clarity on these principles helps with interpreting binomial distribution problems and their real-world implications.
Binomial Distribution Examples
Real-world examples can make the concepts of binomial distribution come alive. A perfect instance is a quality control process where a product either passes or fails inspection (success or failure, respectively), or in medicine, where a treatment either works or does not work on patients. In each of these scenarios, analysts use the number of trials (n) and the probability of a specific outcome (p) to determine the mean and standard deviation of the successes.

For a more direct example from the exercises provided, if a basketball player has a 50% chance of making a free throw (p = 0.5) and they take 100 shots, we expect them to make about 50 shots on average, with a standard deviation that tells us how much variation there might be from this average.

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Most popular questions from this chapter

A shipping company knows that the cost of delivering a small package within 24 hours is \(\$ 14.80 .\) The company charges \(\$ 15.50\) for shipment but guarantees to refund the charge if delivery is not made within 24 hours. If the company fails to deliver only \(2 \%\) of its packages within the 24 -hour period, what is the expected gain per package?

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