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A binomial random variable \(x\) is based on 15 trials with the probability of success equal to 0.2. Find the probability that this variable will take on a value more than 2 standard deviations from the mean.

Short Answer

Expert verified
The exact probability values depend on the calculations in Step 3. All computations involve statistical concepts such as standard deviation, mean and the binomial distribution, as well as summing binomial probabilities.

Step by step solution

01

Compute Mean and Standard Deviation

The mean of a binomial distribution is given by \(\mu=np\), where \(n\) is the number of trials and \(p\) is the probability of success. So, \(\mu=15*0.2=3\). The standard deviation of a binomial distribution is given by \(\sigma=\sqrt{np(1-p)}\). Hence, \(\sigma= \sqrt{15*0.2*0.8}=1.732\).
02

Define Range for the Deviation

We are asked to find the probability that the variable deviates more than 2 standard deviations from the mean. That corresponds to the chances of having less than \(\mu-2\sigma = 3-2*1.732 = -0.464\) or more than \(\mu+2\sigma = 3+2*1.732 = 6.464\) successes. Since the number of successes must be a discrete value, we consider less than 1 success and more than 6 successes.
03

Calculate Probabilities

We use the binomial distribution formula to calculate the probabilities. The formula for binomial probability is \(\frac{n!}{x!(n-x)!}*p^x*(1-p)^{n-x}\) where \(n\) is the number of trials, \(x\) is the number of successes, \(p\) is the probability of success, \(!\) denotes factorial, and \( ^ \) denotes exponentiation. We calculate the probability for \(x=0\) (denotes less than 1) and summation of probabilities for \(x=7\) to \(x=15\) (denotes more than 6). The summation can be obtained by total probability 1 minus the summation from \(x=0\) to \(x=6\). The exact number calculations are complex and can be handled by a calculator or computer software.

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Key Concepts

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

Binomial Random Variable
In statistics, a binomial random variable is a specific type of discrete random variable that counts the number of successes in a fixed number of independent trials of a binomial experiment. For an event that has only two possible outcomes—‘success’ or ‘failure’—we can define a trial, such as flipping a coin or taking a quiz. If we repeat this trial a fixed number of times and the outcome of one trial does not affect the others, we have what is called a binomial scenario.

Each trial in a binomial experiment has the same probability of success, denoted as 'p'. The number of trials is represented by 'n', and the total count of successes is the binomial random variable, denoted as 'x'. In our exercise example, we repeat the trial 15 times and have a success probability of 0.2 for each trial. The binomial random variable 'x' represents the total number of successes out of these 15 trials.
Standard Deviation in Statistics
The standard deviation is a measure of the amount of variation or dispersion of a set of values. In the context of a binomial distribution, the standard deviation provides insights into how spread out the numbers of successes may be over the trials. It calculates the average distance of each data point (or in our case, each possible outcome of successes) from the mean. A high standard deviation means that the actual numbers of successes are spread out widely from the mean, whereas a low standard deviation indicates that the numbers of successes are clustered closely around the mean.

To calculate the standard deviation of a binomial distribution, we use the formula \(\sigma=\sqrt{np(1-p)}\), where 'n' is the number of trials and 'p' is the probability of success. In our example, with 15 trials and a probability of success of 0.2, the standard deviation helps us understand the variability in the value of the binomial random variable 'x'.
Probability of Success
The probability of success in a binomial experiment is a critical component as it directly affects the outcomes and calculations. It is usually denoted as 'p' and represents the chance of a single trial resulting in success. This probability remains constant for all trials within a given binomial experiment and ranges from 0 to 1. Typically, the complementary probability of failure is denoted as '1-p'.

In our example, the probability of success for each trial is 0.2. Knowing this probability allows us to predict the likely outcomes using the binomial distribution. We can calculate the likelihood of achieving a certain number of successes among the fixed number of trials, which is essential for understanding the behavior of the binomial random variable.
Binomial Probability Formula
The binomial probability formula calculates the likelihood of obtaining a specific number of successes in a fixed number of trials in a binomial experiment. This powerful formula is represented as \(P(X=x) = \frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}\), where:\
    \
  • \(n\) is the number of trials\
  • \
  • \(x\) is the number of successes\
  • \
  • \(p\) is the probability of success\
  • \
  • \(!\) denotes factorial\
  • \
  • \( ^ \) denotes exponentiation\
  • \
\
To apply this formula to find specific probabilities, we factor in the total number of ways \(x\) successes can occur among \(n\) trials, multiplied by the probability of success raised to the power of \(x\), and the probability of failure raised to the power of \(n-x\). In our example, we use this formula to calculate the probability of getting 0 successes (indicative of having less than 1 success) and the cumulative probability for getting more than 6 successes by summing individual probabilities from 7 to 15 successes.

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Most popular questions from this chapter

Let \(x\) be a random variable with the following probability distribution: $$\begin{array}{l|cccc}\hline x & 0 & 1 & 2 & 3 \\\P(x) & 0.4 & 0.3 & 0.2 & 0.1 \\\\\hline \end{array}$$ Does \(x\) have a binomial distribution? Justify your answer.

A January 2005 survey of bikers, commissioned by the Progressive Group of Insurance Companies, showed that \(40 \%\) of bikers have body art, such as tattoos and piercings. A group of 10 bikers are in the process of buying motorcycle insurance. a. What is the probability that none of the 10 has any body art? b. What is the probability that exactly 3 have some body art? c. What is the probability that at least 4 have some body art? d. What is the probability that no more than 2 have some body art?

As reported in the chapter opener "USA and Its Automobiles," Americans are in love with the automobile - the majority have more than one vehicle per household. In fact, the national average is 2.28 vehicles per household. The number of vehicles per household in the United States can be described as follows: $$\begin{array}{cc}\text { Vehicles, } \boldsymbol{x} & \boldsymbol{P}(\boldsymbol{x}) \\\\\hline 1 & 0.34 \\\2 & 0.31 \\\3 & 0.22 \\\4 & 0.06 \\\5 & 0.03 \\\6 & 0.02 \\\7 & 0.01 \\\8 \text { or more } & 0.01 \\\\\hline\end{array}$$ a. \(\quad\) Replacing the category "8 or more" with exactly "8," find the mean and standard deviation of the number of vehicles per household in the United States. b. How does the mean calculated in part a correspond to the national average of \(2.28 ?\) c. Explain the effect that replacing the category "8 or more" with "8" has on the mean and standard deviation.

In another germination experiment involving old seed, 50 rows of seeds were planted. The number of seeds germinating in each row were recorded in the following table (each row contained the same number of seeds). $$\begin{array}{cc|cc}\begin{array}{c}\text { Number } \\\\\text { Germinating }\end{array} & \begin{array}{c}\text { Number } \\\\\text { of Rows }\end{array} & \begin{array}{c}\text { Number } \\\\\text { Germinating }\end{array} & \begin{array}{c}\text { Number } \\\\\text { of Rows }\end{array} \\\\\hline 0 & 17 & 3 & 2 \\\1 & 20 & 4 & 1 \\\2 & 10 & 5 \text { or more } & 0 \\\\\hline\end{array}$$ a. What probability distribution (or function) would be helpful in modeling the variable "number of seeds germinating per row"? Justify your choice. b. What information is needed in order to apply the probability distribution you chose in part a? c. Based on the information you do have, what is the highest or lowest rate of germination that you can estimate for these seeds? Explain.

Extended to overtime in game 7 on the road in the 2002 NBA play-offs, the two- time defending champion Los Angeles Lakers did what they do best-thrived when the pressure was at its highest. Both of the Lakers' star players had their chance at the foul line late in overtime. a. With 1: 27 minutes left in overtime and the game tied at \(106-106,\) Shaquille (Shaq) O'Neal was at the line for two free-throw attempts. He has a history of making 0.555 of his free-throw attempts, and during this game, prior to these two shots, he had made 9 of his 13 attempts. Justify the statement "The law of averages was working against him." Both players made both shots, and the series with the Sacramento Kings was over. b. With 0: 06 seconds left in overtime and the game score standing at \(110-106,\) Kobe Bryant was at the line for two free-throw shots. He has a history of making 0.829 of his free throws, and during this game, prior to these two shots, he had made 6 of his 8 attempts. Justify the statement "The law of averages was working for him."

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