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Find the mean and standard deviation for a binomial distribution with these values: a. \(n=1000, p=.3\) b. \(n=400, p=.01\) c. \(n=500, p=.5\) d. \(n=1600, p=.8\)

Short Answer

Expert verified
a. \(n = 1000\), \(p = 0.3\) b. \(n = 400\), \(p = 0.01\) c. \(n = 500\), \(p = 0.5\) d. \(n = 1600\), \(p = 0.8\) Answer: a. \(\mu = 300\), \(\sigma \approx 14.49\) b. \(\mu = 4\), \(\sigma \approx 1.99\) c. \(\mu = 250\), \(\sigma \approx 11.18\) d. \(\mu = 1280\), \(\sigma = 16\)

Step by step solution

01

a. Calculate the mean and standard deviation for \(n=1000\) and \(p=.3\)

Mean: \(\mu = np = 1000 \cdot 0.3 = 300\) Standard Deviation: \(\sigma = \sqrt{np(1-p)} = \sqrt{1000 \cdot 0.3 \cdot (1-0.3)} = \sqrt{210} \approx 14.49\)
02

b. Calculate the mean and standard deviation for \(n=400\) and \(p=.01\)

Mean: \(\mu = np = 400 \cdot 0.01 = 4\) Standard Deviation: \(\sigma = \sqrt{np(1-p)} = \sqrt{400 \cdot 0.01 \cdot (1-0.01)} = \sqrt{3.96} \approx 1.99\)
03

c. Calculate the mean and standard deviation for \(n=500\) and \(p=.5\)

Mean: \(\mu = np = 500 \cdot 0.5 = 250\) Standard Deviation: \(\sigma = \sqrt{np(1-p)} = \sqrt{500 \cdot 0.5 \cdot (1-0.5)} = \sqrt{125} \approx 11.18\)
04

d. Calculate the mean and standard deviation for \(n=1600\) and \(p=.8\)

Mean: \(\mu = np = 1600 \cdot 0.8 = 1280\) Standard Deviation: \(\sigma = \sqrt{np(1-p)} = \sqrt{1600 \cdot 0.8 \cdot (1-0.8)} = \sqrt{256} = 16\) So, the mean and standard deviation of the binomial distributions for the given values are: a. \(\mu = 300\), \(\sigma \approx 14.49\) b. \(\mu = 4\), \(\sigma \approx 1.99\) c. \(\mu = 250\), \(\sigma \approx 11.18\) d. \(\mu = 1280\), \(\sigma = 16\)

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Key Concepts

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

Mean Calculation
In the realm of probability and statistics, the mean of a binomial distribution can be elegantly defined. To calculate the mean, one must remember the formula: \( \mu = np \), where \( n \) represents the number of trials, and \( p \) denotes the probability of success in each trial. This formula provides a snapshot of the expected number of successful outcomes when an experiment is repeated multiple times.

For instance, when you have scenarios such as \( n=1000 \) and \( p=0.3 \), the mean becomes \( 1000 \times 0.3 \), which equals 300. This value represents the average count of successes you expect if you were to conduct 1000 independent tries with a success probability of 0.3 each time.

Key points to consider for calculating mean:
  • Identify the number of trials, \( n \) (total attempts made).
  • Recognize the probability \( p \) (chances of success in each individual trial).
  • Apply these into the formula \( \mu = np \) to find the expected mean.
Standard Deviation Calculation
The standard deviation in a binomial distribution serves a pivotal function. It measures how spread out the successes are from the mean, shedding light on the variability of your data set. The concept shines a spotlight on how far individual elements stray from an average, and for this calculation, we use the formula: \( \sigma = \sqrt{np(1-p)} \).

Let's talk through this concept with an example. Consider the scenario \( n=1000 \) with \( p=0.3 \). By plugging these values into the formula, you get \( \sigma = \sqrt{1000 \times 0.3 \times (1-0.3)} = \sqrt{210} \approx 14.49 \). This approximation tells you that while the mean is at 300, most of your trial's results will hover around this mean within a range of approximately 14.49.

Important reminders for standard deviation:
  • Both \( n \) (trials) and \( p \) (success chance) are essential for calculating \( \sigma \).
  • The formula considers not just successes, but the likelihood of failures through \( (1-p) \).
  • Standard deviation gives insight into the dispersion or spread of possible outcomes.
Probability and Statistics
Understanding binomial distribution isn't complete without a nod towards probability and statistics, as these form the bedrock of the concept. In simple terms, probability measures how likely an event is to occur, while statistics employs these probabilities to make sense of data.

In our binomial distribution scenario, probabilities play a fundamental role in predicting outcomes over a series of trials. These probabilities inform us about whether our expectations, like the mean and variability via standard deviation, align with reality. For instance, when \( p=0.5 \) in a binomial setting with \( n=500 \), the outcomes offer a balanced view where success and failure prospects are evenly weighted.

Binomial distribution relies heavily on two core assumptions:
  • Each trial is independent of the others, meaning past outcomes don't influence future ones.
  • Every trial has only two possible outcomes: success \( (p) \) or failure \( (1-p) \).
By recognizing these elements, one can deftly navigate through studies surrounding probability and statistics, using the mean and standard deviation to produce insightful, data-driven stories.

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Most popular questions from this chapter

A psychiatrist believes that \(80 \%\) of all people who visit doctors have problems of a psychosomatic nature. She decides to select 25 patients at random to test her theory.a. Assuming that the psychiatrist's theory is true, what is the expected value of \(x\), the number of the 25 patients who have psychosomatic problems? b. What is the variance of \(x\), assuming that the theory is true? c. Find \(P(x \leq 14)\). (Use tables and assume that the theory is true.) d. Based on the probability in part \(c\), if only 14 of the 25 sampled had psychosomatic problems, what conclusions would you make about the psychiatrist's theory? Explain.

In a psychology experiment, the researcher designs a maze in which a mouse must choose one of two paths, colored either red or blue, at each of 10 intersections. At the end of the maze, the mouse is given a food reward. The researcher counts the number of times the mouse chooses the red path. If you were the researcher, how would you use this count to decide whether the mouse has any preference for color?

A peony plant with red petals was crossed with another plant having streaky petals. The probability that an offspring from this cross has red flowers is \(.75 .\) Let \(x\) be the number of plants with red petals resulting from ten seeds from this cross that were collected and germinated. a. Does the random variable \(x\) have a binomial distri- bution? If not, why not? If so, what are the values of \(n\) and \(p ?\) b. Find \(P(x \geq 9)\). c. Find \(P(x \leq 1)\). d. Would it be unusual to observe one plant with red petals and the remaining nine plants with streaky petals? If these experimental results actually occurred, what conclusions could you draw?

A balanced coin is tossed three times. Let \(x\) equal the number of heads observed. a. Use the formula for the binomial probability distribution to calculate the probabilities associated with \(x=0,1,2,\) and 3 b. Construct the probability distribution. c. Find the mean and standard deviation of \(x\), using these formulas: $$ \begin{array}{l} \mu=n p \\ \sigma=\sqrt{n p q} \end{array} $$ d. Using the probability distribution in part b, find the fraction of the population measurements lying within one standard deviation of the mean. Repeat for two standard deviations. How do your results agree with Tchebysheff's Theorem and the Empirical Rule?

Let \(x\) be a binomial random variable with \(n=\) 20 and \(p=.1\) a. Calculate \(P(x \leq 4)\) using the binomial formula. b. Calculate \(P(x \leq 4)\) using Table 1 in Appendix I. c. Use the Excel output below to calculate \(P(x \leq 4)\). Compare the results of parts a, b, and c. d. Calculate the mean and standard deviation of the random variable \(x\). e. Use the results of part d to calculate the intervals \(\mu \pm \sigma, \mu \pm 2 \sigma,\) and \(\mu \pm 3 \sigma\). Find the probability that an observation will fall into each of these intervals. f. Are the results of part e consistent with Tchebysheff's Theorem? With the Empirical Rule? Why or why not?

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