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6.89 Suppose that \(20 \%\) of the 10,000 signatures on a certain recall petition are invalid. Would the number of invalid signatures in a sample of size 2000 have (approximately) a binomial distribution? Explain.

Short Answer

Expert verified
In conclusion, the number of invalid signatures in a sample of size 2000 would approximately follow a binomial distribution. This is because, although the trials are not completely independent (since sampling is done without replacement), the sample size is small enough compared to the total population for it to have an approximately negligible effect, and the other conditions for the binomial distribution are satisfied.

Step by step solution

01

Understanding the Problem

We have been provided with the information that 20% (2,000) of the 10,000 signatures on a recall petition are invalid. We want to see if the number of invalid signatures in a sample of size 2000 follows a binomial distribution.
02

Recall Properties of Binomial Distribution

A random variable follows a binomial distribution if the following conditions are met: 1. The number of trials (n) is fixed. 2. Each trial has only two possible outcomes: success (in this case, an invalid signature) and failure (a valid signature). 3. The probability of success (p) is constant for each trial. 4. The trials are independent of each other.
03

Checking the Conditions

Now, let's check if the given scenario satisfies each of these conditions: 1. In the given problem, the number of trials is fixed at n = 2000 signatures. 2. There are only two outcomes: the signature is either invalid (success) or valid (failure). 3. The probability of success (p) is 20%, and the probability of failure (q = 1 - p) is 80%. These probabilities remain constant throughout the trials. 4. Since we are sampling without replacement (as we are not putting back the chosen signature and re-selecting it), the trials are not completely independent. However, the sample size is small compared to the total number of signatures (2000 out of 10,000), and we can apply the Finite Population Correction (FPC) factor and approximate the distribution as a binomial distribution.
04

Conclusion

In conclusion, the number of invalid signatures in a sample of size 2000 would approximately follow a binomial distribution. This is because, although the trials are not completely independent (since sampling is done without replacement), the sample size is small enough compared to the total population for it to have an approximately negligible effect, and the other conditions for the binomial distribution are satisfied.

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Key Concepts

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

Sampling
Sampling is a fundamental concept in statistics. It involves selecting a subset, or sample, from a larger group, known as the population. Here, the population consists of the 10,000 signatures, while we take a sample of 2,000 signatures for further analysis.

This method is vital for gaining information when dealing with large datasets. Sampling allows us to make inferences about the entire population without examining every single item. This saves both time and resources.

There are several methods of sampling, such as:
  • Random Sampling: Every member of the population has an equal chance of being selected.
  • Systematic Sampling: Members are selected at regular intervals.
  • Stratified Sampling: The population is divided into different subgroups, and samples are taken from each.
In the exercise, the sample of 2000 signatures involves random selection, which helps in maintaining the randomness required for statistical analysis.
Probability
Probability is the study of uncertainty and chance, which is essential in determining the likelihood of various outcomes. In our context, we are interested in the probability of selecting an invalid signature from the sample.

The probability is expressed as a number between 0 and 1, where 0 implies an impossible event and 1 indicates certainty. For this exercise:
  • The probability of success (invalid signature) is 20% or 0.2.
  • The probability of failure (valid signature) is 80% or 0.8.
Using these probabilities, we can predict the number of invalid signatures in the sample of 2,000.

Understanding probability is crucial for determining the expected outcomes and making decisions based on statistical data. It enables us to evaluate the feasibility of certain hypotheses, like the assumption of a binomial distribution in this case.
Finite Population Correction
The Finite Population Correction (FPC) is a useful adjustment made when sampling from a finite population without replacement. It helps in ensuring more accurate results when the sample size is not negligible compared to the population size.

In our example with the 10,000 signatures, we sample 2,000. While this number might not be very small, it is still only a fifth of the total, which keeps the sampling influence relatively minor. However, technically, such sampling without replacement makes the trials less than perfectly independent.

Normally, to account for this lack of independence, we apply the FPC, which slightly adjusts our calculations compared to cases involving infinite populations, or where the sample size is very small. The correction factor is given by:\[ \text{FPC} = \sqrt{\frac{N-n}{N-1}} \] where \( N \) is the population size, and \( n \) is the sample size.

Applying FPC allows the approximation of the distribution as binomial, as it ensures that small deviations due to sampling without replacement do not significantly affect the analysis.

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Most popular questions from this chapter

Suppose a playlist on an MP3 music player consisting of 100 songs includes 8 by a particular artist. Suppose that songs are played by selecting a song at random (with replacement) from the playlist. The random variable \(x\) represents the number of songs until a song by this artist is played. a. Explain why the probability distribution of \(x\) is not binomial. b. Find the following probabilities. (Hint: See Example \(6.31 .\) ) i. \(p(4)\) ii. \(P(x \leq 4)\) iii. \(P(x>4)\) iv. \(P(x \geq 4)\) c. Interpret each of the probabilities in Part (b) and explain the difference between them.

A soft-drink machine dispenses only regular Coke and Diet Coke. Sixty percent of all purchases from this machine are diet drinks. The machine currently has 10 cans of each type. If 15 customers want to purchase drinks before the machine is restocked, what is the probability that each of the 15 is able to purchase the type of drink desired?

A machine that cuts corks for wine bottles operates in such a way that the distribution of the diameter of the corks produced is well approximated by a normal distribution with mean \(3 \mathrm{~cm}\) and standard deviation \(0.1 \mathrm{~cm} .\) The specifications call for corks with diameters between 2.9 and \(3.1 \mathrm{~cm}\). A cork not meeting the specifications is considered defective. (A cork that is too small leaks and causes the wine to deteriorate; a cork that is too large doesn't fit in the bottle.) What proportion of corks produced by this machine are defective?

Suppose that the amount of time spent by a statistical consultant with a client at their first meeting is a random variable that has a normal distribution with a mean value of 60 minutes and a standard deviation of 10 minutes. a. What is the probability that more than 45 minutes is spent at the first meeting? b. What amount of time is exceeded by only \(10 \%\) of all clients at a first meeting?

The light bulbs used to provide exterior lighting for a large office building have an average lifetime of 700 hours. If lifetime is approximately normally distributed with a standard deviation of 50 hours, how often should all the bulbs be replaced so that no more than \(20 \%\) of the bulbs will have already burned out?

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