Problem 34 - Basic Statistics for Business and Economics

Question

A study of long distance phone calls made from the corporate offices of the Pepsi Bottling Group, Inc., in Somers, New York, showed the calls follow the normal distribution. The mean length of time per call was 4.2 minutes and the standard deviation was 0.60 minutes. a. What fraction of the calls last between 4.2 and 5 minutes? b. What fraction of the calls last more than 5 minutes? c. What fraction of the calls last between 5 and 6 minutes? d. What fraction of the calls last between 4 and 6 minutes? e. As part of her report to the president, the Director of Communications would like to report the length of the longest (in duration) 4 percent of the calls. What is this time?

Step-by-Step Solution

Answer

a. 40.82% of calls last between 4.2 and 5 minutes. b. 9.18% last more than 5 minutes. c. 9.05% last between 5 and 6 minutes. d. 62.80% last between 4 and 6 minutes. e. The longest 4% of calls last more than 5.25 minutes.

Step 1: Understanding the Problem

We are tasked with finding the fraction of phone calls that last within certain time durations based on a normal distribution with a mean of 4.2 minutes and a standard deviation of 0.60 minutes. We will use the standard normal distribution (z-distribution) to find probabilities.

Step 2: Calculating Z-scores

Convert the time durations into standard z-scores using the formula: \[ z = \frac{X - \mu}{\sigma} \]where \( X \) is the value, \( \mu = 4.2 \) minutes is the mean, and \( \sigma = 0.60 \) minutes is the standard deviation.

Step 3: Fraction of Calls Between 4.2 and 5 Minutes

Find the z-scores for 4.2 and 5 minutes: - For 4.2 minutes: \( z = \frac{4.2 - 4.2}{0.60} = 0 \)- For 5 minutes: \( z = \frac{5 - 4.2}{0.60} \approx 1.33 \)Look up these z-scores in the z-table. The probability between these is \( P(Z < 1.33) - P(Z < 0) \approx 0.9082 - 0.5 = 0.4082 \).

Step 4: Fraction of Calls More Than 5 Minutes

Using the z-score for 5 minutes, find the probability: \[ P(Z > 1.33) = 1 - P(Z < 1.33) \approx 1 - 0.9082 = 0.0918 \]

Step 5: Fraction of Calls Between 5 and 6 Minutes

Find the z-score for 6 minutes: \[ z = \frac{6 - 4.2}{0.60} = 3 \]Probability for z less than 3 is \( P(Z < 3) \approx 0.9987 \). Therefore, \[ P(5 < X < 6) = P(Z < 3) - P(Z < 1.33) \approx 0.9987 - 0.9082 = 0.0905 \]

Step 6: Fraction of Calls Between 4 and 6 Minutes

Find the z-score for 4 minutes:\[ z = \frac{4 - 4.2}{0.60} = -0.33 \]Look up the z-scores in the z-table: \[ P(-0.33 < Z < 3) \approx P(Z < 3) - P(Z < -0.33) = 0.9987 - 0.3707 = 0.6280 \]

Step 7: Longest 4% of Calls

To find the length of the longest 4% of calls, look for the z-score corresponding to the 96th percentile (since 100% - 4% = 96%). From the z-table, this z-score is approximately 1.75. Calculate:\[ X = \mu + z \cdot \sigma = 4.2 + 1.75 \, \times \, 0.60 \approx 5.25 \]

Key Concepts

z-scores

A z-score is a crucial concept when dealing with normal distributions. It essentially tells us how many standard deviations away a particular data point is from the mean. To calculate a z-score, use the formula: \[ z = \frac{X - \mu}{\sigma} \]Where:

• \(X\) is the observed value
• \(\mu\) is the mean of the data set
• \(\sigma\) is the standard deviation

A z-score of 0 means the data point is exactly at the mean. Positive z-scores represent values above the mean, while negative z-scores represent values below the mean. By converting data into z-scores, we can look up probabilities in a standard normal distribution table, enabling us to determine how common or rare a data point is within a distribution.
For example, in our exercise, converting 5 minutes to a z-score tells us how much longer than average these calls last compared to a typical call.

standard deviation

Standard deviation is a measure of how spread out the numbers in a data set are. In simpler terms, it tells us how much the values in a set typically differ from the mean (average). A smaller standard deviation means the values are closer to the mean, while a larger one indicates a wider spread.The formula for standard deviation is:\[ \sigma = \sqrt{\frac{\sum (X - \mu)^2}{N}} \]Where:

• \(X\) represents each value in the dataset
• \(\mu\) is the mean
• \(N\) is the number of data points

In the context of our problem, knowing the standard deviation (0.60 minutes, in this case) allows us to calculate z-scores, which are used for probability calculations and understanding the distribution of phone call durations.

probability calculations

Probability calculations are fundamental for interpreting normal distributions. These calculations help us determine the likelihood of a data point falling within a certain range. Once data points are converted into z-scores, these can be used alongside z-tables or calculators to find probabilities. For instance, when you know the z-score related to a particular phone call duration, you can look it up in a z-table to determine the probability of such a call lasting that long. In our exercise, we calculated that the probability of a call lasting between 5 and 6 minutes is approximately 0.0905, or around 9.05%, by looking at the cumulative probabilities for the calculated z-scores. This assists in making informed decisions based on past data, which is particularly beneficial for business and statistical analysis.

percentiles in statistics

Percentiles indicate the relative standing of a value within a data set. If you know a value's percentile rank, you understand what proportion of the data is below it. Percentiles are especially common in educational testing and statistical report presentations.To find percentiles in a normal distribution, you often use z-scores. For example, if you're interested in the value below which 96% of phone call durations fall, you find this by identifying the z-score that corresponds to the 96th percentile. In the exercise provided, we wanted the maximum duration for the longest 4% of calls, which means we looked for the 96th percentile. By using the z-score for this percentile, which is approximately 1.75, we could calculate the corresponding call duration using the formula:\[ X = \mu + z \cdot \sigma \]Understanding percentiles helps in setting benchmarks and interpreting relative performance across different datasets.