Problem 14 - Basic Statistics for Business and Economics with Student

Question

The United States Postal Service reports 95 percent of first class mail within the same city is delivered within two days of the time of mailing. Six letters are randomly sent to different locations. a. What is the probability that all six arrive within two days? b. What is the probability that exactly five arrive within two days? c. Find the mean number of letters that will arrive within two days. d. Compute the variance and standard deviation of the number that will arrive within two days.

Step-by-Step Solution

Answer

a) 0.735; b) 0.233; c) 5.7; d) Variance: 0.285, Standard deviation: 0.534.

Step 1: Define Success Probability and Sample Size

The probability of success (a letter arriving within two days) is given as 95%, which can be expressed as \( p = 0.95 \). The number of trials, which in this case is the number of letters, is \( n = 6 \). We will use these values throughout the solution to apply the binomial probability formula.

Step 2: Calculate Probability for All Six Letters

To find the probability that all six letters arrive within two days, use the formula \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \). Here, \( k = 6 \):\[ P(X = 6) = \binom{6}{6} (0.95)^6 (0.05)^0 \]Since \( \binom{6}{6} = 1 \) and \( (0.05)^0 = 1 \), the calculation becomes:\[ P(X = 6) = (0.95)^6 \approx 0.735 \]

Step 3: Calculate Probability for Exactly Five Letters

To calculate the probability that exactly five letters arrive within two days, set \( k = 5 \):\[ P(X = 5) = \binom{6}{5} (0.95)^5 (0.05)^1 \]Calculate \( \binom{6}{5} = 6 \), then:\[ P(X = 5) = 6 \times (0.95)^5 \times 0.05 \approx 0.233 \]

Step 4: Determine the Mean Number of Arrivals

The mean number of letters arriving within two days is calculated by the formula for the mean of a binomial distribution: \( \mu = n \times p \).\[ \mu = 6 \times 0.95 = 5.7 \]

Step 5: Compute the Variance and Standard Deviation

The variance of a binomial distribution is given by \( \sigma^2 = n \times p \times (1-p) \):\[ \sigma^2 = 6 \times 0.95 \times 0.05 = 0.285 \]The standard deviation is the square root of the variance:\[ \sigma = \sqrt{0.285} \approx 0.534 \]

Key Concepts

Probability Theory

Binomial distribution is a key concept in probability theory. It allows us to model the likelihood of a certain number of successes in a fixed number of trials. Here, a 'success' is defined as an event that has a specified outcome. In the context of our exercise, a success is a letter arriving within two days.

The binomial distribution requires two parameters:

• The probability of a single success, denoted by \( p \). In our case, \( p = 0.95 \), as reported by the postal service.
• The sample size or number of trials, denoted by \( n \). For the exercise, \( n = 6 \) since six letters are mailed.

The probability mass function (PMF) of a binomial distribution is used to calculate probabilities of various events. It's expressed by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] This formula tells us the probability of achieving exactly \( k \) successes in \( n \) trials.

Mean and Variance

The mean and variance are fundamental statistical measures that provide deeper insights into the binomial distribution. Mean represents the average outcome we expect over a large number of trials. In a binomial setting, the mean \( \mu \) is calculated by multiplying the number of trials by the probability of success: \[ \mu = n \times p \] For our postal example, this yields a mean of \( 5.7 \), indicating that, on average, out of six letters, around 5.7 letters arrive within two days.

Variance \( \sigma^2 \) gives us an idea of how much the outcomes can differ from the mean. It's calculated as: \[ \sigma^2 = n \times p \times (1-p) \] For this scenario, the variance comes out to be \( 0.285 \). The standard deviation, \( \sigma \), which is the square root of the variance, provides a more comprehensible measure of this variability: \[ \sigma = \sqrt{0.285} \approx 0.534 \] These measures help in understanding not only what the average number of successes will be but also how much variation we can expect around this average.

Statistical Calculations

Performing statistical calculations involves applying mathematical formulas to determine specific probabilities and statistical measures. In a binomial distribution, key calculations include precise probabilities for different outcomes and the assessment of mean and variance.

1. **Calculating Exact Probabilities** : Using the binomial probability formula, we can find the likelihood that exactly \( k \) successes occur. - For all six letters arriving on time, we calculate \( P(X = 6) \), resulting in approximately 0.735. - For exactly five letters, \( P(X = 5) \) is approximately 0.233. These precise probabilities help us understand the most likely outcomes.
2. **Assessing Expectation and Spread** : Mean and variance are derived to represent the expected number of successes and their variability: - The mean \( (5.7) \) shows the overall expectation out of the six mailed letters. - Variance \( (0.285) \) and standard deviation \( (0.534) \) assess the expected spread around the mean.
These calculations are vital for making informed predictions and understanding possible outcomes in probability scenarios.