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Response Rates for Random Digit Dialing. When an opinion poll uses random digit dialing to select respondents for polls, the response rate is approximately \(10 \%\) for people contacted by cell phone. You watch a pollster dial 20 cell numbers using random digit dialing. a. What is the probability that exactly two calls yield a response? b. What is the probability that at most two calls yield a response? c. What is the probability that at least two calls yield a response? d. What is the probability that fewer than two calls yields a response? e. What is the probability that more than two calls yield a response?

Short Answer

Expert verified
a. 0.2852 b. 0.6778 c. 0.3222 d. 0.6778 e. 0.3222

Step by step solution

01

Understanding the Problem

The problem involves a binomial distribution since each call results in a response (success) or no response (failure). The probability of success \( p = 0.1 \), and we have 20 trials (calls). We are calculating several probabilities based on this distribution.
02

Find the Probability of Exactly Two Responses

To find the probability of exactly two responses, use the binomial probability formula: \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]Substitute \( n = 20 \), \( k = 2 \), and \( p = 0.1 \): \[P(X = 2) = \binom{20}{2} (0.1)^2 (0.9)^{18}\] Calculate this to find the desired probability.
03

Calculate At Most Two Responses

To find the probability of at most two responses, sum the probabilities of 0, 1, and 2 responses:\[P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)\]Calculate each term using the binomial formula, sum them to get the total probability.
04

Calculate At Least Two Responses

To find the probability of at least two responses, use the complement rule:\[P(X \geq 2) = 1 - P(X < 2)\]We already know \( P(X < 2) = P(X = 0) + P(X = 1) \) from Step 3. Subtract this from 1 to get the probability of at least two responses.
05

Calculate Fewer Than Two Responses

Using the results from Step 3 directly,\[P(X < 2) = P(X = 0) + P(X = 1)\]This gives the probability that fewer than two calls yield a response.
06

Calculate More Than Two Responses

To find the probability of more than two responses, the complement of at most two responses is used:\[P(X > 2) = 1 - P(X \leq 2)\]Subtract the sum found in Step 3 from 1 to get the probability of more than two responses.

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Key Concepts

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

Probability Calculations
Probability calculations help in understanding the likelihood of different outcomes in a given scenario. In our exercise, we use the binomial distribution, which is perfect for situations with two possible outcomes, such as success and failure. Here, success means a phone response, and failure means no response. The formula for the binomial probability is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] - \( n \) represents the number of trials (or phone calls in our case). - \( k \) is the number of successful trials (responses). - \( p \) is the probability of success for each trial (10% here).These calculations help us find the probability of getting exactly two responses, at most two, at least two, more than two, and fewer than two, giving us a complete picture of the possible outcomes. Every calculation is a step toward these findings, offering insights into how likely events are in random scenarios.
Response Rate
The response rate is a crucial aspect in surveys and opinion polls. With random digit dialing, this exercise illustrates a typical example where the response rate is given as 10%. - A response rate of 10% indicates that, on average, just 10 out of every 100 dialed numbers provide a response. - This understanding helps us set expectations on survey outcomes and design statistical models to forecast future results. The challenge is enhanced when multiple trials are involved, as the response rate directly influences our success probability in binomial calculations. Understanding response rates aids in managing and interpreting data, particularly in research, and ensuring the efficiency of surveys.
Random Digit Dialing
Random Digit Dialing (RDD) is a sampling method widely used in surveys, where telephone numbers are randomly generated and dialed. This method aims to achieve a representative sample of the population, minimizing selection bias. With RDD, each call can either result in a response (success) or no response (failure), perfectly fitting the binomial distribution model. It's crucial in probabilistic models because it ensures variability and randomness, key components in statistical analysis. The randomness of RDD is an essential factor in predicting responses and understanding probability calculations in surveys.
Complement Rule
The complement rule is a fundamental principle in probability that helps find the probability of at least or more than a certain number of successes. It states that the probability of an event happening is 1 minus the probability of it not happening. In the exercise:- To find the probability of at least two responses, we calculate the complement of having fewer than two responses: \[ P(X \geq 2) = 1 - P(X < 2) \] By understanding and using the complement rule, we efficiently find probabilities in situations with complex outcome combinations, without recalculating all possibilities. This rule simplifies calculations, saving time and reducing errors in statistical analysis.

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Most popular questions from this chapter

Response Rates for Random Digit Dialing. When an opinion poll uses random digit dialing to select respondents for polls, the response rate is approximately \(10 \%\) for people contacted by cell phone. You watch a pollster dial 20 cell numbers many times using random digit dialing. a. What is the mean number of calls that yield a response? b. What is the standard deviation \(\sigma\) of the count of calls that yield a response? c. Suppose that the probability of getting a response were \(p=0.05\). How does this new \(p\) affect the standard deviation? What would be the standard deviation if \(p=0.01\) ? What does your work show about the behavior of the standard deviation of a binomial distribution as the probability of success gets closer to zero?

False Positives in Testing for HIV. A rapid test for the presence in the blood of antibodies to HIV, the virus that causes AIDS, gives a positive result with probability about \(0.004\) when a person who is free of HIV antibodies is tested. A clinic test s 1000 people who are all free of HIV antibodies. a. What is the distribution of the number of positive tests? b. What is the mean number of positive tests? c. You cannot safely use the Normal approximation for this distribution. Explain why.

Larry reads that half of all super jumbo eggs contain double yolks. So he always buys super jumbo eggs and uses two whenever he cooks. If eggs do or don't contain two yolks independently of each other, the number of eggs with double yolks when Larry uses two chosen at random has the distribution a. binomial with \(n=2\) and \(p=1 / 2\). b. binomial with \(n=2\) and \(p=1 / 3\). c. binomial with \(n=3\) and \(p=1 / 2\).

Estimating \(\pi\) from Random Numbers. Kenyon College student Eric Newman used basic geometry to evaluate software random number generators as part of a summer research project. He generated 2000 independent random points \((X, Y)\) in the unit square. (That is, \(X\) and \(Y\) are independent random numbers between 0 and 1 , each having the density function illustrated in Eigure \(12.5\) (page 284). The probability that \((X, Y)\) falls in any region within the unit square is the area of the region.) 13 a. Sketch the unit square, the region of possible values for the point \((X, Y)\). b. The set of points \((X, Y)\) where \(X^{2}+Y^{2}<1\) describes a circle of radius 1. Add this circle to your sketch in part (a) and label the intersection of the two regions \(A\). c. Let \(T\) be the total number of the 2000 points that fall into the region \(A\). T follows a binomial distribution. Identify \(n\) and \(p\). (Hint: Recall that the area of a circle is \(\pi r^{2}\).) d. What are the mean and standard deviation of T? e. Explain how Eric used a random number generator and the facts given here to estimate \(\pi\).

Is This Coin Balanced? While he was a prisoner of war during World War II, John Kerrich tossed a coin 10,000 times. He got 5067 heads. If the coin is perfectly balanced, the probability of a head is \(0.5\). Is there reason to think that Kerrich's coin was not balanced? To answer this question, find the probability that tossing a balanced coin 10,000 times would give a count of heads at least this far from 5000 (that is, at least 5067 heads or no more than 4933 heads).

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