Problem 14 - Basic Statistics for Business & Economics

Question

The U.S. Postal Service reports \(95 \%\) of first-class mail within the same city is delivered within 2 days of the time of mailing. Six letters are randomly sent to different locations. a. What is the probability that all six arrive within 2 days? b. What is the probability that exactly five arrive within 2 days? c. Find the mean number of letters that will arrive within 2 days. d. Compute the variance and standard deviation of the number that will arrive within 2 days.

Step-by-Step Solution

Answer

(a) 0.7351; (b) 0.2322; (c) 5.7; (d) Variance: 0.285, SD: 0.5347.

Step 1: Determine the Probability Model

Since we are dealing with letters arriving within 2 days, which can either happen or not, this scenario follows a Binomial Distribution. This is because each letter represents an independent Bernoulli trial where the probability of success (arrival within 2 days) is given as \( p = 0.95 \), and the number of trials (letters) is 6.

Step 2: Calculate Probability That All Six Letters Arrive Within 2 Days

To find the probability that all six letters arrive within 2 days, we use the binomial probability formula: \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \), where \( n = 6 \), \( k = 6 \), and \( p = 0.95 \). Substitute these values into the formula to get \( P(X = 6) = \binom{6}{6} (0.95)^6 (0.05)^0 = (0.95)^6 \). Calculate this to get the probability value.

Step 3: Compute Specific Probability: All Six Arrive

Calculate \((0.95)^6 \). Using a calculator, this equals approximately \(0.7351\). Thus, the probability that all six letters arrive within 2 days is \(0.7351\) or 73.51\%.

Step 4: Calculate Probability That Exactly Five Letters Arrive Within 2 Days

Use the binomial probability formula to find \( P(X = 5) \): \( P(X = 5) = \binom{6}{5} (0.95)^5 (0.05)^1 \). Here, \( \binom{6}{5} = 6 \), so calculate \( 6 \times (0.95)^5 \times (0.05) \).

Step 5: Compute Specific Probability: Exactly Five Arrive

Calculate \((0.95)^5 = 0.7738\) and then \(6 \times 0.7738 \times 0.05\). This equals \(6 \times 0.7738 \times 0.05 = 0.2322\). Thus, the probability that exactly five letters arrive within 2 days is \(0.2322\) or 23.22\%.

Step 6: Find the Mean Number of Letters Arriving Within 2 Days

For a binomial distribution, the mean \(\mu\) is given by \(\mu = n \times p\). Here, \( n = 6 \) and \( p = 0.95 \). So, the mean is \( 6 \times 0.95 = 5.7 \). This means, on average, 5.7 out of 6 letters are expected to arrive within 2 days.

Step 7: Calculate the Variance

The variance \(\sigma^2\) for a binomial distribution is given by \(\sigma^2 = n \times p \times (1-p)\). Using \( n = 6 \), \( p = 0.95 \), and \(1-p = 0.05\), calculate: \(\sigma^2 = 6 \times 0.95 \times 0.05 = 0.285\).

Step 8: Determine the Standard Deviation

The standard deviation is the square root of the variance. Therefore, \(\sigma = \sqrt{0.285} \approx 0.5347\). This gives the spread of delivery times for the letters.

Key Concepts

Probability

In the world of statistics, probability is a crucial concept that helps us quantify the likelihood of different outcomes. In our mail delivery problem, we are dealing with a binomial distribution. Each letter's delivery within 2 days is a simple yes or no, i.e., success or failure. The probability of success in each trial (each letter) is given as 0.95.

A binomial distribution is used when there are a fixed number of independent trials (like the 6 letters), each with the same probability of success. We calculate specific probabilities using the binomial probability formula:

• The probability that all six letters arrive within 2 days is calculated as \[P(X = 6) = \binom{6}{6} (0.95)^6 (0.05)^0 = (0.95)^6 \]. This equals approximately 0.7351, meaning there's a 73.51% chance all will arrive on time.
• If we want precisely five letters to succeed in arrival on time, the formula is \[P(X = 5) = \binom{6}{5} (0.95)^5 (0.05)^1\]. This probability is approximately 0.2322 or 23.22%.

These calculations show how likely different outcomes are, providing a complete picture of what we can expect on average.

Mean and Variance

Another key aspect of the binomial distribution involves understanding the mean and variance.

The **mean** (often represented as \( \mu \)) provides the average expected outcome. For our mail scenario, the mean is calculated using the formula:

• \(\mu = n \times p\)

where \( n = 6 \) and \( p = 0.95 \). Hence, the mean number of letters expected to arrive on time is 5.7.This suggests that, on average, about 5 or 6 letters will arrive within the 2-day period.

The **variance** measures how much the results will vary from this mean on average. It's calculated as:

• \(\sigma^2 = n \times p \times (1-p)\)

Substituting the values, we have \( 6 \times 0.95 \times 0.05 = 0.285 \). This tells us about the distribution of numbers around the mean.

Understanding these concepts helps us anticipate variability and set expectations about delivery performance.

Standard Deviation

Standard deviation is another measure closely related to variance. While variance provides a measure of how data is spread around the mean, standard deviation gives us a direct and interpretable figure. It represents the average amount each number in a distribution differs from the mean.

Calculating the standard deviation for our binomially distributed mail scenario involves taking the square root of the variance. If the variance \(\sigma^2\) is 0.285, then the standard deviation \(\sigma\) is calculated as:

• \(\sigma = \sqrt{0.285} \approx 0.5347\)

This result shows that the number of letters that might deviate from the mean (5.7 letters) is roughly 0.53.

In practical terms, understanding the standard deviation allows us to expect some variability, but still provides confidence in the average delivery scenario. It's a valuable tool for predicting outcomes in statistical practices and forms the foundation of understanding data distribution patterns.