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Chapter 5: Problem 89

A binomial random variable has a mean equal to 200 and a standard deviation of \(10 .\) Find the values of \(n\) and \(p.\)

Short Answer

Expert verified
The values for n and p are 400 and 0.5, respectively.

Step by step solution

01

Derive the Mean and Variance Equations

First, we can set up equations based on the formulas for the mean and variance of a binomial distribution. The mean, denoted as 渭, is calculated as in the formula \( n*p \), and the variance, denoted as 蟽虏, is calculated as in the formula \(n*p*(1-p)\). Given that the mean is 200 and standard deviation (which is the square root of the variance) is 10, we can write down the following equations from these formulas: \n Equation 1: \(n*p = 200\), \n Equation 2: \(n*p*(1 - p) = 100\).
02

Solve the System of Equations

Now, there are two equations to solve for n and p. One way to do this is to express n from Equation 1 in terms of p, and then substitute that into Equation 2. \n From equation 1 we get: \(n = 200/p\), \n Substituting \(n = 200/p\) in equation 2, we get: \(200 - 200*p = 100\). Solving this equation for p gives \(p = 0.5\).
03

Find the Value of n

Once we know p, we can substitute p = 0.5 into Equation 1 to find n. Solving for n in equation 1 gives us \(n = 200 / p = 200 / 0.5 = 400\).

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Key Concepts

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

Mean of Binomial Distribution
The mean of a binomial distribution represents the average number of successes in a series of independent trials of a binomial experiment. The mathematical formula for the mean (碌) of a binomial distribution with n trials and a probability p of success in each trial is \[\begin{equation} \mu = n \cdot p \end{equation}\]Where:
  • n represents the number of trials,
  • p represents the probability of success on a single trial.
In the context of the given exercise, with a mean indicated as 200, the equation establishes a direct relationship between the number of trials and the probability of each trial's success. Understanding this formula is essential in solving for the values of 'n' and 'p' when they are unknown.
Standard Deviation
Standard deviation in the context of a binomial distribution measures the amount of variability or dispersion of a set of values from the mean. The lower the standard deviation, the closer the data points (in this case, the number of successes) are to the mean. Conversely, a high standard deviation implies that the data points are spread out over a wider range of values.To calculate the standard deviation (蟽) for a binomial distribution, we take the square root of the variance (蟽虏). The formula for the variance of a binomial distribution is given by:\[\begin{equation} \sigma^2 = n \cdot p \cdot (1 - p) \end{equation}\]The given exercise states that the standard deviation is 10, which means that when we apply this formula, the variance, 蟽虏, is 100. This variance is an intrinsic part of the distribution's spread, helping to determine 'n' and 'p' by working together with the mean.
Binomial Distribution Equations
With the mean and standard deviation formulas in hand, we construct a system of equations to identify the unknown parameters of a binomial distribution, namely the number of trials (n) and the probability of success (p). In the provided exercise, we derived two equations:\[\begin{equation} n \cdot p = 200 \end{equation}\] (Equation 1, Mean)and\[\begin{equation} n \cdot p \cdot (1 - p) = 100 \end{equation}\] (Equation 2, Variance)These equations are pivotal as they allow us to solve for 'n' and 'p' by substituting one equation into the other. Once we solve for 'p', 'n' can be easily found as demonstrated in the step-by-step solution. This systematic approach of using equations particular to binomial distributions serves as a bedrock for solving various statistical problems involving binomial random variables.

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Most popular questions from this chapter

Evaluate each of the following. a. \(4 !\) b. \(7 !\) c. \(0 !\) d. \(\frac{6 !}{2 !}\) e. \(\frac{5 !}{2 ! 3 !}$$ f. \)\frac{6 !}{4 !(6-4) !}$$ g. \((0.3)^{4}$$ h. \)\left(\begin{array}{l}7 \\\3 \end{array}\right)\( i. \)(\begin{array}{l}5 \\ 2\end{array}\right)\( j . \)(\begin{array}{l}3 \\ 0\end{array}\right)\( k. \)\left(\begin{array}{l}4 \\ 1\end{array}\right)(0.2)^{1}(0.8)^{3}\( l. \)\left(\begin{array}{l}5 \\ 0\end{array}\right)(0.3)^{0}(0.7)^{5}$

The increase in Internet usage over the past few years has been phenomenal, as demonstrated by the February 2004 report from the Pew Internet \& American Life Project. The survey of Americans 65 or older (about 8 million adults) reported that \(22 \%\) have access to the Internet. By contrast, \(58 \%\) of 50 to 64 -year-olds, \(75 \%\) of 30 - to 49 -year-olds, and \(77 \%\) of 18 - to 29 -year-olds currently go online. Suppose that 50 adults in each age group are to be interviewed. a. What is the probability that "have Internet access" is the response of 10 to 20 adults in the 65 or older group? b. What is the probability that "have Internet access" is the response of 30 to 40 adults in the 50 - to 64-year-old group? c. What is the probability that "have Internet access" is the response of 30 to 40 adults in the 30 - to 49-year-old group? d. What is the probability that "have Internet access" is the response of 30 to 40 adults in the 18 - to 29-year-old group? e. Why are the answers for parts a and d nearly the same? Explain. f. What effect did the various values of \(p\) have on the probabilities? Explain.

Show that each of the following is true for any values of \(n\) and \(k\). Use two specific sets of values for \(n\) and \(k\) to show that each is true.. $$\text { a. } \quad\left(\begin{array}{l}n \\\0\end{array}\right)=1 \text { and }\left(\begin{array}{l}n \\\n\end{array}\right)=1$$ $$\text { b. }\left(\begin{array}{l}n \\\1\end{array}\right)=n \text { and }\left(\begin{array}{c}n \\\n-1 \end{array}\right)=n \quad \text { c. }\left(\begin{array}{c}n \\\k\end{array}\right)=\left(\begin{array}{c}n \\\n-k\end{array}\right)$$

In a USA Today Snapshot (June 1,2009), the following statistics were reported on the number of hours of sleep that adults get. $$\begin{aligned}&\begin{array}{cc}\text { Number of Hours } & \text { Percentage } \\\\\hline 5 \text { or less } & 12 \% \\\6 & 29 \% \\\7 & 37 \% \\\8 \text { or more } & 24 \% \\\\\hline\end{array}\\\&\begin{array}{l} \text { Source: Strategy One survey for Tempur-Pedic of } \\\1004 \text { adults in April }\end{array} \end{aligned}$$ a. Are there any other values that the number of hours can attain? b. Explain why the total of the percentages is not \(100 \%\) c. Is this a discrete probability distribution? Is it a probability distribution? Explain.

Draw a tree diagram picturing a binomial experiment of four trials.
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