Problem 23 - Basic Statistics for Business & Economics

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 \(70 \%\) 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) Expect 10.5 resolved, SD 1.77; b) P(10 resolved) ≈ 0.2061; c) P(10 or 11 resolved) ≈ 0.4569; d) P(more than 10 resolved) ≈ 0.6509.

Step 1: Determine Expected Value (a)

The expected value in a binomial distribution can be calculated using the formula \( E(X) = n \times p \), where \( n \) is the number of trials, and \( p \) is the probability of success on each trial. For this problem, \( n = 15 \) and \( p = 0.7 \). Thus, \( E(X) = 15 \times 0.7 = 10.5 \). You expect 10.5 problems to be resolved today.

Step 2: Calculate Standard Deviation (a)

The standard deviation for a binomial distribution is calculated using the formula \( \sigma = \sqrt{n \times p \times (1-p)} \). With \( n = 15 \) and \( p = 0.7 \), the standard deviation is \( \sigma = \sqrt{15 \times 0.7 \times 0.3} = \sqrt{3.15} \approx 1.77 \).

Step 3: Probability of Exactly 10 Resolved (b)

To find the probability of exactly 10 successes in a binomial distribution, use the binomial probability formula: \[ P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} \].Substitute \( n = 15 \), \( k = 10 \), and \( p = 0.7 \) into the equation to get:\[ P(X = 10) = \binom{15}{10} \cdot (0.7)^{10} \cdot (0.3)^{5} \approx 0.2061 \].The probability that exactly 10 problems are resolved today is approximately 0.2061.

Step 4: Probability of 10 or 11 Resolved (c)

We need to find \( P(X = 10) + P(X = 11) \). We already calculated \( P(X = 10) \approx 0.2061 \).Now calculate \( P(X = 11) \) using: \[ P(X = 11) = \binom{15}{11} \cdot (0.7)^{11} \cdot (0.3)^{4} \approx 0.2508 \].So, \( P(X = 10 \text{ or } 11) = P(X = 10) + P(X = 11) \approx 0.2061 + 0.2508 = 0.4569 \).

Step 5: Probability of More than 10 Resolved (d)

First calculate the probability for 11, 12, 13, 14, and 15 problems resolved and sum them up:- \( P(X = 11) \approx 0.2508 \) (already calculated).- \( P(X = 12) = \binom{15}{12} \cdot (0.7)^{12} \cdot (0.3)^{3} \approx 0.2311 \).- \( P(X = 13) = \binom{15}{13} \cdot (0.7)^{13} \cdot (0.3)^{2} \approx 0.1240 \).- \( P(X = 14) = \binom{15}{14} \cdot (0.7)^{14} \cdot (0.3)^{1} \approx 0.0403 \).- \( P(X = 15) = \binom{15}{15} \cdot (0.7)^{15} \cdot (0.3)^{0} \approx 0.0047 \).Sum these probabilities: \[ P(X > 10) \approx 0.2508 + 0.2311 + 0.1240 + 0.0403 + 0.0047 = 0.6509 \].

Key Concepts

Expected Value

The expected value is a fundamental concept in probability and statistics. It provides a measure of the central tendency of a random variable in a probability distribution. When dealing with a binomial distribution, you can find the expected value using a straightforward formula:

• The formula is: \( E(X) = n \times p \).
• Here, \( n \) is the total number of trials, and \( p \) is the probability of success on each trial.

In our given problem, we are told that the Georgetown Telephone Company (GTC) resolves customer complaints on the same day 70% of the time. Today, there are 15 complaints. Thus, to calculate the expected number of resolved complaints, we apply the formula:

• Substitute \( n = 15 \) and \( p = 0.7 \) into the formula.
• The expected value is therefore \( 15 \times 0.7 = 10.5 \).

This means on average, GTC will resolve approximately ten and a half customer complaints today.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. For a binomial distribution, it tells us how much the number of successes in trials is expected to deviate from the expected value. The formula for finding the standard deviation in a binomial distribution is:

• \( \sigma = \sqrt{n \times p \times (1-p)} \)
• Here, \( n \) is the number of trials, \( p \) is the probability of success, and \( 1-p \) represents the probability of failure.

Using the GTC example, where \( n = 15 \) and \( p = 0.7 \), we calculate it as:

• Substitute the values to get \( \sigma = \sqrt{15 \times 0.7 \times 0.3} \).
• This results in approximately \( \sigma = \sqrt{3.15} \approx 1.77 \).

The calculated standard deviation of approximately 1.77 indicates the extent to which the number of successfully resolved complaints can vary from the expected average.

Probability

Probability helps us understand the likelihood of different outcomes in uncertain situations. In the context of binomial distributions, probability tells us how likely a certain number of successes is over a specific number of trials. For example, one might be interested in the probability of exactly 10 out of 15 complaints being resolved on a particular day by GTC. To calculate this, we use the binomial probability formula.

• Each probability calculated represents the likelihood that a certain number of successes will occur.
• In particular, calculating \( P(X = 10) \) involves determining the likelihood of having exactly ten successes.

Ultimately, understanding probability allows us to make more informed predictions about activities based on historical data, such as GTC's complaint resolutions.

Binomial Probability Formula

The binomial probability formula is used to calculate the probability of a given number of successes in a set number of trials, each with the same probability of success. The basic formula is:

• \[ P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} \]
• Here, \( n \) represents the total number of trials, \( p \) the probability of success, and \( k \) the specific number of successes you are interested in.

For instance, to find the probability of exactly 10 out of 15 GTC complaints being resolved, substitute \( n = 15 \), \( p = 0.7 \), and \( k = 10 \) into the formula:

• \[ P(X = 10) = \binom{15}{10} \cdot (0.7)^{10} \cdot (0.3)^{5} \approx 0.2061 \]

This calculation process can be repeated for any specific scenario within the parameters of the problem to derive meaningful probabilities and inform decision-making.