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The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports they can resolve customer problems the same day they are reported in 70 percent of the cases. Suppose the 15 cases reported today are representative of all complaints. a. How many of the problems would you expect to be resolved today? What is the standard deviation? b. What is the probability 10 of the problems can be resolved today? c. What is the probability 10 or 11 of the problems can be resolved today? d. What is the probability more than 10 of the problems can be resolved today?

Short Answer

Expert verified
Expected: 10.5 cases; Std Dev: ≈1.78; \(P(X=10) ≈ 0.2061\); \(P(X=10\,or\,11) ≈ 0.4396\); \(P(X>10) ≈ 0.5124\).

Step by step solution

01

Identify Given Information

The problem involves a binomial distribution where the number of trials, \( n \), is 15 (15 cases reported today) and the probability of success, \( p \), is 0.7 (70% of cases resolved on the same day).
02

Calculate Expected Value

The expected value (mean) for a binomial distribution is given by the formula: \( E(X) = n \cdot p \). So, \( E(X) = 15 \cdot 0.7 = 10.5 \). This means, on average, 10.5 cases will be resolved today.
03

Calculate Standard Deviation

The standard deviation for a binomial distribution is calculated using the formula: \( \sigma = \sqrt{n \cdot p \cdot (1-p)} \). Substitute the values: \( \sigma = \sqrt{15 \cdot 0.7 \cdot 0.3} \approx 1.78 \).
04

Probability of Resolving 10 Cases

For a binomial random variable, the probability of exactly \( k \) successes is given by: \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \). Here, \( k = 10 \), \( n = 15 \), \( p = 0.7 \). Calculate: \( P(X = 10) = \binom{15}{10} (0.7)^{10} (0.3)^5 \approx 0.2061 \).
05

Probability of Resolving 10 or 11 Cases

We need to find \( P(X = 10) + P(X = 11) \). \( P(X = 11) = \binom{15}{11} (0.7)^{11} (0.3)^4 \approx 0.2335 \). Therefore, \( P(X = 10 \, \text{or} \, 11) = 0.2061 + 0.2335 = 0.4396 \).
06

Probability of Resolving More Than 10 Cases

To find the probability of resolving more than 10 cases, calculate \( P(X > 10) \), which is \( 1 - P(X \leq 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) \). Calculating each term, \( P(X > 10) \approx 1 - 0.4876 = 0.5124 \).

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Key Concepts

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

Expected Value
The concept of expected value is a central component of probability and statistics, especially when dealing with random variables like in our binomial distribution case. The expected value can be thought of as the long-term average you would expect to see if you performed the same experiment many times. For a binomial distribution, where we're dealing with a set number of trials () and a consistent probability of success (), the expected value is calculated using the formula: \[ E(X) = n \cdot p \]In our specific exercise, we need to find out how many customer issues Georgetown Telephone Company can typically resolve in one day. Given that there are 15 cases and the resolution probability for each is 0.7, the expected value tells us that on average, 10.5 cases will be resolved. This result helps the company anticipate workload and resource needs for improving efficiency.
Standard Deviation
Standard deviation provides insight into the variability or spread of data in a probability distribution. For a binomial distribution, it offers information on how much the data points differ from the expected value. It is important because, while the expected value gives an average, the standard deviation tells us how spread out the results typically are.The formula for the standard deviation in a binomial distribution is:\[ \sigma = \sqrt{n \cdot p \cdot (1-p)} \]In our case study, with 15 trials and a 70% success rate, we plug these into the formula to get a standard deviation of approximately 1.78. This means that the number of cases resolved daily will typically vary by about 1.78 cases from the expected 10.5. This variability accounts for the natural fluctuations in day-to-day operations.
Probability Calculation
Probability calculation in binomial distributions often involves finding the probability of achieving a certain number of successes from a series of trials. Here, the formula we use is:\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]The binomial coefficient \( \binom{n}{k} \) represents the number of ways \( k \) successes can occur out of \( n \) trials. In our example, the probability that exactly 10 cases are resolved is calculated by substituting the respective values, which results in approximately 0.2061.Furthermore, calculating for multiple scenarios such as 10 or 11 cases resolved involves adding individual probabilities: \[ P(X = 10) + P(X = 11) \approx 0.2061 + 0.2335 = 0.4396 \]These calculations allow businesses to predict specific outcomes and make informed decisions.
Binomial Random Variable
A binomial random variable is a type of discrete random variable that models the number of successes in a given number of Bernoulli trials. Each trial is independent and results in a success with probability \( p \) or a failure with probability \( 1-p \). The binomial distribution is quite prevalent because it models several real-world processes and experiments efficiently.In our case with the Georgetown Telephone Company, the variable of interest is the number of customer issues resolved in one day, which can either be resolved (success) or unresolved (failure) based on the reported probability of 0.7. Understanding the binomial random variable helps in assessing probabilities of different outcomes like resolving more than 10 cases, calculated as:\[ P(X > 10) = 1 - P(X \leq 10) = P(X = 11) + P(X = 12) + ... + P(X = 15) \]This type of analysis provides insights into operational strategies based on expected performance.

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