Problem 23 - Basic Statistics for Business and Economics with Student

Question

The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it 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?

Step-by-Step Solution

Answer

a. Expected: 10.5, Std Dev: ~1.77. b. \( P(X = 10) \). c. \( P(X = 10) + P(X = 11) \). d. \( 1 - P(X \leq 10) \).

Step 1: Understand the Scenario

We are given a probability problem involving a binomial distribution. GTC resolves issues on the same day with a probability of 0.7. We have 15 reported cases, and we need to use these parameters to answer the questions.

Step 2: Define the Variables

Identify the binomial distribution parameters: \( n = 15 \) (number of trials), \( p = 0.7 \) (probability of success). We will be interested in various probabilities concerning the number of cases resolved the same day (\( X \)), which follows the binomial distribution \( X \sim B(15, 0.7) \).

Step 3: Solve Part a - Expected Value and Standard Deviation

The expected value of a binomial distribution is given by \( E(X) = np \). So, \( E(X) = 15 \times 0.7 = 10.5 \). The standard deviation is calculated using \( \sigma = \sqrt{np(1-p)} \). Thus, \( \sigma = \sqrt{15 \times 0.7 \times 0.3} = \sqrt{3.15} \approx 1.77 \).

Step 4: Solve Part b - Probability of Exactly 10 Resolved

The probability of exactly 10 successes (resolved problems) is found using the binomial probability formula: \[ P(X = 10) = \binom{15}{10} (0.7)^{10} (0.3)^{5} \]Calculate using a calculator or software to find \( P(X = 10) \).

Step 5: Solve Part c - Probability of 10 or 11 Resolved

We need the probability of the events \( X = 10 \) or \( X = 11 \). This can be found by summing the separate probabilities:\[ P(X = 10 \, \text{or} \, X = 11) = P(X = 10) + P(X = 11) \]Calculate \( P(X = 11) \) using \( \binom{15}{11} (0.7)^{11} (0.3)^{4} \) and add to \( P(X = 10) \).

Step 6: Solve Part d - Probability More Than 10 Resolved

The probability that more than 10 problems are resolved is \( P(X > 10) \). This is calculated as:\[ P(X > 10) = 1 - P(X \leq 10) \]Calculate \( P(X \leq 10) \) which is the cumulative probability for \( X = 0, 1, \ldots, 10 \), then subtract from 1.

Key Concepts

Probability

In the context of a binomial distribution, probability plays a vital role in predicting outcomes. Probability refers to the likelihood of an event happening. For our exercise, GTC has a probability of 0.7 (or 70%) of resolving issues within the same day. This probability enables us to predict various outcomes using the characteristics of a binomial distribution.

• Each event (or trial) in a binomial experiment can land in one of two categories: success or failure. Here, a "success" means a problem is resolved on the same day.
• The overall probability remains constant for each trial.
• We focus on 15 trials, meaning we consider 15 reported issues.
• To calculate specific probabilities, such as 10 problems being resolved, we use the binomial probability formula.

The use of this formula allows us to determine the probability of exactly 10 problems getting resolved today. This is a core part of understanding binomial probability functions.

Expected Value

Expected Value, often denoted as \( E(X) \), represents the average outcome we anticipate based on probability over a large number of trials. It's equivalent to the weighted average of all possible outcomes.
In our example, the expected value is crucial as it provides insight into the number of problems GTC is likely to resolve on any given day. To calculate this, we use the formula for expected value in a binomial distribution: \( E(X) = np \).

• \( n \) is the number of trials, which here is 15 problem reports.
• \( p \) is the probability of resolving an issue the same day, which is 0.7.

By multiplying these values, we find \( E(X) = 15 \times 0.7 = 10.5 \). This means, on average, GTC can expect to resolve 10.5 problems daily if such a large sample is consistently evaluated. This expected value provides a general expectation, yet the exact number can vary in smaller or individual samples.

Standard Deviation

Standard deviation in the context of binomial distribution provides a measure of the variation or spread of the outcome probabilities around the expected value. It helps us understand how much the actual resolved problems can deviate from the average value we calculated.
The formula for standard deviation \( \sigma \) in a binomial distribution is given by \( \sigma = \sqrt{np(1-p)} \). It captures the variability in resolving issues.

• We calculate \( np(1-p) = 15 \times 0.7 \times 0.3 \) which simplifies to 3.15.
• The standard deviation is then \( \sqrt{3.15} \approx 1.77 \).

This means that typically, the number of problems resolved will vary by about 1.77 from the expected 10.5. Understanding standard deviation is crucial, as it gives context to how consistent the results are compared to what is expected.

Cumulative Probability

Cumulative probability helps us understand the likelihood that a variable falls within a specified range. It sums the probabilities of all possible outcomes up to a certain point.
For instance, finding the probability that more than 10 problems can be resolved involves calculating cumulative probabilities.

• First, we determine \( P(X \leq 10) \), which represents resolving up to 10 issues.
• This involves adding the probabilities for each possible outcome from 0 through 10.
• To find \( P(X > 10) \), we use \( 1 - P(X \leq 10) \).

By evaluating these cumulative probabilities, we can understand the broader picture of how many problems are likely to be resolved. This concept is powerful in calculating probabilities for ranges of outcomes rather than a single specific outcome.