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Chapter 3: Problem 187

The probability that your call to a service line is answered in less than 30 seconds is \(0.75 .\) Assume that your calls are independent. (a) If you call 10 times, what is the probability that exactly nine of your calls are answered within 30 seconds? (b) If you call 20 times, what is the probability that at least 16 calls are answered in less than 30 seconds? (c) If you call 20 times, what is the mean number of calls that are answered in less than 30 seconds?

Short Answer

Expert verified
(a) 0.3336; (b) Use software for exact probability; (c) 15 calls.

Step by step solution

01

Identify the type of problem

This exercise involves determining probabilities based on a series of independent events where each event has two possible outcomes: answered in less than 30 seconds or not. This is a typical problem suitable for a binomial distribution.
02

Define Binomial Distribution Parameters

For a binomial distribution, the parameters are the number of trials \( n \) and the probability of success \( p \). Here, \( n = 10 \) or \( 20 \) (depending on the part of the question) and \( p = 0.75 \).
03

Step 3a: Calculate Probability for Part (a)

To find the probability that exactly nine out of ten calls are answered within 30 seconds, use the binomial probability formula: \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}.\] For \( n = 10 \), \( k = 9 \), and \( p = 0.75 \): \[P(X = 9) = \binom{10}{9} (0.75)^9 (0.25)^1.\] Calculate this step-by-step.
04

Step 4a: Calculate Combination for Part (a)

Calculate the binomial coefficient for this scenario: \( \binom{10}{9} = 10 \).
05

Step 5a: Compute Probabilities for Part (a)

Now calculate \( (0.75)^9 \) which is approximately \( 0.1335 \) and \( (0.25)^1 = 0.25 \). Multiply all the calculated values: \[P(X = 9) = 10 \times 0.1335 \times 0.25 \approx 0.3336.\]
06

Step 3b: Find Probability for Part (b)

To find the probability that at least 16 out of 20 calls are answered within 30 seconds, sum the probabilities for each case from 16 to 20 using the binomial probability formula. \[P(X \geq 16) = \sum_{k=16}^{20} \binom{20}{k} (0.75)^k (0.25)^{20-k}.\] Calculate each term separately.
07

Step 4b: Calculate Individual Probabilities for Part (b)

Calculate each probability separately for \( k = 16 \) to \( k = 20 \) using: \( P(X = k) = \binom{20}{k} (0.75)^k (0.25)^{20-k} \). Sum them to find \( P(X \geq 16) \). This may require computational software or a calculator specializing in statistics.
08

Step 3c: Calculate Mean for Part (c)

The mean of a binomial distribution is given by \( \mu = n \times p \). For 20 calls where \( p = 0.75 \):\[\mu = 20 \times 0.75 = 15.\] Thus, the mean number of calls answered in less than 30 seconds is 15.

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Key Concepts

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

Probability Calculations
Probability calculations in binomial distributions can be tackled with a systematic approach. The given problem tells us about a scenario where calls are either answered in less than 30 seconds or not. This problem involves the calculation of probabilities for independent trials, making the binomial distribution a suitable model.

To compute the probability of a specific number of successes (in this case, calls answered quickly) in a fixed number of trials, you apply the binomial probability formula:
  • \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\] where:
  • \(n\) is the number of trials,
  • \(k\) is the number of successful trials,
  • \(p\) is the probability of success in each trial.
This formula is used to calculate the probability of exactly k successes out of n trials. A binomial coefficient, \(\binom{n}{k}\), also known as "n choose k", determines the number of ways to choose k successes from n trials.
Independent Events
In statistics, understanding independent events is crucial when dealing with probability calculations. Independent events are those whose outcomes do not affect one another. For example, each call in this problem is independent. The result of one call does not change the probability of the next call being answered within 30 seconds.

Independence plays a vital role in using the binomial distribution, as it allows taking multiple trials without expecting the probabilities to shift through those trials. Specifically:
  • An event is independent if changing the outcome does not affect the likelihood of other events.
  • The consistency of probabilities is key in sequences of independent trials.
For instance, just because one of your calls was answered quickly does not inherently alter the chance of the next call being answered swiftly.
When events are independent, probabilities can be multiplied across trials to determine the overall likelihood of multiple successes, as applied in the original problem.
Mean of Distribution
The mean or expected value of a binomial distribution provides an average number of successes one can expect over a specified number of trials. Calculating the mean is straightforward for a binomial distribution:

  • The mean is given by the formula: \(\mu = n \times p\)
  • Where \(n\) represents the number of trials, and \(p\) is the probability of success per trial.
For our example, if you make 20 calls with a 0.75 probability that each call is answered under 30 seconds, the mean \(\mu = 20 \times 0.75 = 15\).
Thus, on average, you can expect 15 calls to be answered in the desired time frame.
The mean offers a central point of the distribution and acts as a predictive measure, giving insight into the typical outcome of repeated trials.

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Most popular questions from this chapter

The number of messages that arrive at a Web site is a Poisson random variable with a mean of five messages per hour. (a) What is the probability that five messages are received in 1.0 hour? (b) What is the probability that 10 messages are received in 1.5 hours? (c) What is the probability that fewer than two messages are received in 0.5 hour?

Show that the probability density function of a negative binomial random variable equals the probability density function of a geometric random variable when \(r=1\). Show that the formulas for the mean and variance of a negative binomial random variable equal the corresponding results for a geometric random variable when \(r=1\).

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