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Chapter 14: Problem 2

Working Cell Numbers. When an opinion poll selects cell phone numbers at random to dial, the cell phone exchange is first selected, and then random digits are added to form a complete telephone number (see Example 8.5). When using this procedure to generate random cell phone numbers, approximately \(55 \%\) of the cell numbers generated correspond to working numbers. You watch a pollster dial cell numbers that have been selected in this manner. \(X\) is the number of calls dialed before the third working cell number is reached.

Short Answer

Expert verified
The probability of needing exactly 6 calls to reach the third working number is 0.151.

Step by step solution

01

Understand the Problem

The problem involves randomly dialing cell phone numbers to reach working numbers. We want to find the probability distribution of the number of calls needed to reach the third working number. This is a negative binomial distribution problem where each call results in a success (working number) with a probability of 0.55.
02

Define the Parameters

Let's define the success probability as \( p = 0.55 \), since 55% of the numbers are working. We are asked for the number of trials (calls) needed to reach the third success (working number), so \( k = 3 \).
03

Use the Negative Binomial Formula

The negative binomial distribution gives the probability of \( x \) trials needed to get \( k \) successes, with a success probability \( p \). The formula is: \[ P(X = x) = \binom{x-1}{k-1} p^k (1-p)^{x-k} \]For this problem, \( k = 3 \) and \( p = 0.55 \).
04

Calculate the Probability for a Specific Example

As an example, let's say we want to find the probability that exactly 6 calls are needed (meaning the third working number is the 6th call). Plug the values into the formula:\[ P(X=6) = \binom{5}{2} (0.55)^3 (0.45)^{3} \]Calculate \( \binom{5}{2} = 10 \), and evaluate the powers of 0.55 and 0.45, then multiply.
05

Solve the Expression

Calculate as follows:- \( (0.55)^3 = 0.166375 \)- \( (0.45)^3 = 0.091125 \)- \( \binom{5}{2} = 10 \)So\[ P(X=6) = 10 \cdot 0.166375 \cdot 0.091125 = 0.1510434375 \]
06

Conclusion

This calculation shows that the probability of needing exactly 6 calls to reach the third working number is approximately 0.151, dependent on the given success rate.

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Key Concepts

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

Probability Distribution
In statistics, a probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes for an experiment. In the case of dialing random cell phone numbers, each call represents an occurrence that can be classified as either reaching a working number, which is a success, or not, which is a failure.
  • Different types of probability distributions can describe random processes, such as the binomial, normal, or negative binomial distributions.
  • The choice of distribution depends on the specific characteristics and constraints of the problem.
  • For the scenario given, the negative binomial distribution is used to model the number of phone calls needed until a set number of working (success) calls are achieved.
The negative binomial distribution specifically accounts for sequences of Bernoulli trials, where there are two possible outcomes for each trial (success or failure). In this sense, it expands on the binomial distribution by focusing on the number of trials needed to achieve a specified number of successes.
Success Probability
The success probability, denoted as \( p \, \), measures the likelihood of a successful outcome in each independent trial. It is crucial in determining how probability weights are assigned across different scenarios.
  • In this exercise, the success probability is 0.55, meaning there is a 55% chance each dialed number will be working.
  • This probability directly influences the calculations of the negative binomial probability.
  • The greater the success probability, the fewer trials may be needed to achieve a set number of successes.
For example, in our scenario, if the probability of reaching a working number were larger, fewer calls would be expected on average before reaching the third working number. On the other hand, if the probability were smaller, more calls would likely be required.
Binomial Coefficient
The binomial coefficient, symbolized as \( \binom{x}{y} \, \), is a central component of the binomial theorem. It represents the number of ways to choose \( y \, \) successes from \( x \, \) trials, regardless of order.
  • This coefficient is crucial when calculating probabilities involving binomial distributions, including negative binomial scenarios.
  • It is computed using the formula: \[ \binom{x}{y} = \frac{x!}{y!(x-y)!} \]
  • For example, in the problem where we calculate the probability for 6 calls to reach the third success, the binomial coefficient is \( \binom{5}{2} = 10 \).
This value represents the number of distinct ways to arrange 3 working numbers within 6 total calls. Understanding how to find and apply binomial coefficients is essential for working with binomial distributions.
Random Sampling
Random sampling is a fundamental statistical process used to select a subset of a population to represent the entire group. It aims to eliminate bias in the selection process so that each sample member has an equal opportunity for selection.
  • In this exercise, random sampling is used to dial cell phone numbers randomly to determine if they are working numbers.
  • This helps ensure that the sample of numbers dialed is representative of the entire population of potential cell phone numbers.
  • Such sampling is key in surveys and polls to obtain valid and reliable outcomes.
Random sampling underlies the reliability of the results derived from probability distributions. Without proper random sampling, the inferences drawn from statistical analyses could be skewed or misleading.

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Most popular questions from this chapter

Roulette-betting on red. A roulette wheel has 38 slots, numbered 0,00 , and 1 to 36 . The slots 0 and 00 are colored green, 18 of the others are red, and 18 are black. The dealer spins the wheel and, at the same time, rolls a small ball along the wheel in the opposite direction. The wheel is carefully balanced so that the ball is equally likely to land in any slot when the wheel slows. Gamblers can bet on various combinations of numbers and colors. (a) If you bet on "red," you win if the ball lands in a red slot. What is the probability of winning with a bet on red in a single play of roulette? (b) You decide to play roulette four times, each time betting on red. What is the distribution of \(X\), the number of times you win? (c) If you bet the same amount on each play and win on exactly two of the four plays, then you will "break even." What is the probability that you will break even? (d) If you win on fewer than two of the four plays, then you will lose money. What is the probability that you will lose money?

Using Benford's Law. According to Benford's law (Example 12.7, page 2B7) the probability that the first digit of the amount of a randomly chosen invoice is a 1 or a 2 is \(0.477\). You examine 90 invoices from a vendor and find that 29 have first digits 1 or 2 . If Benford's law holds, the count of 1 s and \(2 s\) will have the binomial distribution with \(n=90\) and \(p=0.477\). Too few \(1 \mathrm{~s}\) and \(2 \mathrm{~s}\) suggests fraud. What is the approximate probability of 29 or fewer 15 and \(2 \mathrm{~s}\) if the invoices follow Benford's law? Do you suspect that the invoice amounts are not genuine?

Is this coin balanced? While he was a prisoner of war during World War II, 14.36 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).

Checking for Survey Errors. One way of checking the effect of undercoverage, nonresponse, and other sources of error in a sample survey is to compare the sample with known facts about the population. About \(22 \%\) of the Canadian population is first generation, that is, they were bom outside Canada. \({ }^{6}\) The number \(X\) of first-generation Canadians in random samples of 1500 persons should therefore vary with the binomial ( \(n=1500, p=0.22\) ) distribution. (a) What are the mean and standard deviation of \(X\) ? (b) Use the Nomal approximation to find the probability that the sample will contain between 340 and 390 first-generation Canadians. Be sure to check that you can safely use the approximation.

Multiple-choice tests. Here is a simple probability model for multiple-choince tests. Suppose each student has probability p of correctly answering a question chosen at random from a universe of possible quest矛ons. (A strong student has a higher \(p\) than a weak student.) Answers to different questions are independent. (a) Stacey is a good student for whom \(p=0.75\). Use the Normal approximation to find the probability that Stacey scores between \(70 \%\) and \(80 \%\) on a 100 -question test. (b) If the test contains 250 questions, what is the probability that Stacey will score between \(70 \%\) and \(80 \%\) ? You see that Stacey's score on the longer test is more likely to be close to her "true score."
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