/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 determine whether the given rand... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

determine whether the given random variable has a binomial distribution. Justify your answer. Sowing seeds Seed Depot advertises that its new flower seeds have an \(85 \%\) chance of germinating (growing). Suppose that the company's claim is true. Judy gets a packet with 20 randomly selected new flower seeds from Seed Depot and plants them in her garden. Let \(X=\) the number of seeds that germinate.

Short Answer

Expert verified
The random variable X has a binomial distribution because it meets all binomial conditions.

Step by step solution

01

Define a Binomial Distribution

A binomial distribution is defined by certain conditions: there must be a fixed number of trials, each trial has only two possible outcomes (success or failure), the probability of success must be the same in each trial, and the trials must be independent.
02

Identify Trials and Outcomes

In this scenario, Judy plants 20 seeds, which represent the fixed number of trials. Each seed can either germinate (success) or not germinate (failure), satisfying the requirement of two possible outcomes.
03

Check Probability Consistency

The problem states that each seed has an 85% chance of germinating, so the probability of success (germination) for each seed is 0.85. This probability is consistent for each trial, fulfilling the condition of a constant probability of success.
04

Confirm Independence

Assume the germination of each seed is independent of the others. This means the success of one seed's germination does not affect another's, which is a standard assumption in binomial distribution unless stated otherwise.
05

Conclusion

Given that the scenario satisfies all the conditions: a fixed number of trials (20 seeds), two possible outcomes per trial, constant probability of success (0.85), and independent trials, the random variable X, the number of seeds that germinate, follows a binomial distribution.

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Key Concepts

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

Probability
Probability is a fundamental concept in statistics and mathematics that measures how likely an event is to occur. In the context of binomial distribution, probability refers to the likelihood that a certain number of successes will happen over a set number of trials. Here, when discussing the germination of seeds, we're concerned with the probability that each seed will sprout.

In our problem concerning Judy's seeds, the probability of a seed germinating is given as 85%, or expressed as a decimal, 0.85. This means that out of any given seed, there's an 85% chance it will successfully germinate if the company's claim holds true. The concept of probability helps us predict the outcomes of random events.

In a binomial distribution, these probabilities are fixed and do not change from one trial to the next, which is crucial for determining whether the situation fits the criteria. By using this fixed probability, we can model how likely different numbers of successes, such as seeds germinating, are across multiple trials.
Random Variable
A random variable is a variable that represents the outcome of a random process. It quantifies the results of experiments or events that are inherently uncertain. In binomial distributions, a random variable often counts the number of successes in a given number of independent trials.

In the case of Judy and her flower seeds, the random variable is defined as \(X\), where \(X\) represents the number of seeds that germinate. Since each seed might or might not grow, \(X\) could range anywhere from 0 to 20, depending on how many seeds actually germinate.

Random variables are essential because they allow us to use statistical tools to determine probabilities and expected values. In the context of binomial distributions, we focus on discrete random variables, which have a countable number of possible outcomes.
Independent Trials
Independent trials are a key aspect of the binomial distribution. Trials are considered independent if the result of one trial does not affect the outcome of another. This means that each trial's results are separate and do not influence other trials.

For Judy's garden, independent trials would mean that whether or not one seed germinates doesn't change the chance of another seed germinating. Each seed operates as its own trial, with an independent probability of success or failure.

Independence is crucial in probability calculations because it simplifies the estimation process. If trials were not independent, calculating probabilities would require considering additional complex interactions. In theoretical and practical scenarios, assuming independence allows us to apply the binomial distribution more easily, provided that we have no evidence to the contrary.

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Most popular questions from this chapter

The Tri-State Pick 3 Most states and Canadian provinces have government- sponsored lotteries. Here is a simple lottery wager, from the Tri-State Pick 3 game that New Hampshire shares with Maine and Vermont. You choose a number with 3 digits from 0 to \(9 ;\) the state chooses a three-digit winning number at random and pays you \(\$ 500\) if your number is chosen. Because there are 1000 numbers with three digits, you have probability \(1 / 1000\) of winning. Taking \(X\) to be the amount your ticket pays you, the probability distribution of \(X\) is $$ \begin{array}{lcc} \hline \text { Payoff: } & \$ 0 & \$ 500 \\ \text { Probability: } & 0.999 & 0.001 \\ \hline \end{array} $$ (a) Show that the mean and standard deviation of \(X\) are $$ \mu_{X}=\$ 0.50 \text { and } \sigma_{X}=\$ 15.80 $$ (b) If you buy a Pick 3 ticket, your winnings are \(W=X-1\), because it costs \(\$ 1\) to play. Find the mean and standard deviation of \(W\). Interpret each of these values in context.

"Large Counts condition To use a Normal distribution to approximate binomial probabilities, why do we require that both \(n p\) and \(n(1-p)\) be at least \(10 ?\)

Rainy days Imagine that we randomly select a day from the past 10 years. Let \(X\) be the recorded rainfall on this date at the airport in Orlando, Florida, and \(Y\) be the recorded rainfall on this date at Disney World just outside Orlando. Suppose that you know the means \(\mu_{X}\) and \(\mu_{Y}\) and the variances \(\sigma_{X}^{2}\) and \(\sigma_{Y}^{2}\) of both variables. (a) Is it reasonable to take the mean of the total rainfall \(X+Y\) to be \(\mu_{X}+\mu_{Y} ?\) Explain your answer. (b) Is it reasonable to take the variance of the total rainfall to be \(\sigma_{\mathrm{X}}^{2}+\sigma_{\mathrm{Y}}^{2}\) ? Explain your answer.

Working out Choose a person aged 19 to 25 years at random and ask, "In the past seven days, how many times did you go to an exercise or fitness center or work out?" Call the response \(Y\) for short. Based on a large sample survey, here is a probability model for the answer you will get: \({ }^{8}\) $$ \begin{array}{lcccccccc} \hline \text { Days: } & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \text { Probability: } & 0.68 & 0.05 & 0.07 & 0.08 & 0.05 & 0.04 & 0.01 & 0.02 \\\ \hline \end{array} $$ (a) Show that this is a legitimate probability distribution. (b) Make a histogram of the probability distribution. Describe what you see. (c) Describe the event \(Y<7\) in words. What is \(P(Y<7) ?\) (d) Express the event "worked out at least once" in terms of \(Y\). What is the probability of this event?

Exercises 27 to 29 refer to the following setting. Choose an American household at random and let the random variable \(X\) be the number of cars (including \(\mathrm{SUVs}\) and light trucks) they own. Here is the probability model if we ignore the few households that own more than 5 cars: \begin{tabular}{lcccccc} \hline Number of cars \(X:\) & 0 & 1 & 2 & 3 & 4 & 5 \\ Probability: & 0.09 & 0.36 & 0.35 & 0.13 & 0.05 & 0.02 \\ \hline \end{tabular} 27\. What's the expected number of cars in a randomly selected American household? (a) 1.00 (b) 1.75 (c) 1.84 (d) 2.00 (e) 2.50

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