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A study of long-distance phone calls made from the corporate offices of Pepsi Bottling Group, Inc., in Somers, New York, revealed the length of the calls, in minutes, follows the normal probability 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?

Short Answer

Expert verified
Use Z-scores to find probabilities and longest call duration.

Step by step solution

01

Understand the Given Problem

We are given a normal distribution of the length of phone calls, with a mean of \( \mu = 4.2 \) minutes and a standard deviation of \( \sigma = 0.6 \) minutes. Various probabilities related to this distribution need to be calculated.
02

Convert to Standard Normal Distribution (Z-Scores)

To solve the problems, we convert call durations to Z-scores, which are calculated by the formula \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the call length.
03

Calculate Fraction of Calls Between 4.2 and 5 Minutes

The Z-score for 4.2 minutes is \( Z = \frac{4.2 - 4.2}{0.6} = 0 \) and for 5 minutes is \( Z = \frac{5 - 4.2}{0.6} = 1.33 \). Using Z-tables, find the probability for these Z-scores: \( P(Z < 1.33) \). The fraction of calls between 4.2 and 5 minutes is \( P(Z < 1.33) - P(Z < 0) \).
04

Calculate Fraction of Calls More than 5 Minutes

From the previous step, we have \( P(Z < 1.33) \). The probability more than 5 minutes is \( 1 - P(Z < 1.33) \).
05

Calculate Fraction of Calls Between 5 and 6 Minutes

Find the Z-score for 6 minutes: \( Z = \frac{6 - 4.2}{0.6} = 3 \). The fraction of calls between 5 and 6 minutes is \( P(1.33 < Z < 3) = P(Z < 3) - P(Z < 1.33) \).
06

Calculate Fraction of Calls Between 4 and 6 Minutes

Calculate the Z-score for 4 minutes: \( Z = \frac{4 - 4.2}{0.6} = -0.33 \). The fraction between 4 and 6 minutes is \( P(-0.33 < Z < 3) = P(Z < 3) - P(Z < -0.33) \).
07

Identify the Longest 4% of Calls

This requires finding the Z-score where the cumulative probability from left to right is 0.96 (since the upper tail is 4%). Find this Z-score from the table and then convert it to time using \( X = Z \times \sigma + \mu \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Z-Scores
Z-scores are a crucial concept in the world of statistics, especially when dealing with normal distribution. They allow us to determine how far away a specific data point is from the mean, measured in terms of standard deviations. This standardization enables the comparison of data points from different normal distributions.

To calculate a Z-score, use the formula:
  • \( Z = \frac{X - \mu}{\sigma} \)
Here, \(X\) is the actual value for which you are calculating the Z-score, \(\mu\) is the mean of the distribution, and \(\sigma\) is the standard deviation. A Z-score of zero indicates that the data point is exactly at the mean. For instance, in the problem with phone call durations from the corporate office, a call lasting exactly 4.2 minutes results in a Z-score of 0. This helps compare it easily against other call durations across the normal distribution pattern.
Probability Distribution
A probability distribution describes how the probabilities of different outcomes are distributed. In our case, the phone call durations are normally distributed. This is the bell-shaped curve that is symmetric around the mean, representing how data points are spread around the central point.

The normal distribution is key in statistics and often assumed in many natural phenomena. It is characterized by the mean (central tendency) and the standard deviation (how spread out the values are). The total area under the curve equals to 1, reflecting the probability that a data point falls within a certain range. For example, with the telephone calls that had a mean of 4.2 minutes and a standard deviation of 0.6, the probability distribution allows us to quantify the likelihood of a call being of a particular length.
Mean and Standard Deviation
The mean and standard deviation are fundamental in summarizing normal distributions. The mean \(\mu\) refers to the average value of a dataset, providing a central location of the data points. In our telephone call scenario, the mean duration is 4.2 minutes, acting as the center of the distribution.

Standard deviation, denoted by \(\sigma\), measures the dispersion of the dataset relative to the mean. A small standard deviation indicates that data points are close to the mean, while a larger one signifies that data points are spread out over a wider range. In the phone call dataset, a standard deviation of 0.6 minutes shows the extent of variation from the mean. This is vital for calculating Z-scores and determining probabilities in a normal distribution.
Cumulative Probability
Cumulative probability refers to the likelihood of a random variable being less than or equal to a certain value. It is the integral of the probability distribution function over a specific range.

In normal distributions, cumulative probability helps determine the probability between two values or beyond a certain point. For example, to find the fraction of phone calls exceeding 5 minutes, you calculate the cumulative probability for Z-scores less than 1.33 and subtract it from 1. This tells you how likely it is for a call to last more than 5 minutes. Similarly, to find the top 4% longest calls, you use the cumulative probability to locate the Z-score that corresponds to 96% and then convert that Z-score back to the actual time using the mean and standard deviation.

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Most popular questions from this chapter

The mean of a normal probability distribution is 400 pounds. The standard deviation is 10 pounds. a. What is the area between 415 pounds and the mean of 400 pounds? b. What is the area between the mean and 395 pounds? c. What is the probability of selecting a value at random and discovering that it has a value of less than 395 pounds?

A normal population has a mean of 12.2 and a standard deviation of 2.5 a. Compute the \(z\) value associated with 14.3 . b. What proportion of the population is between 12.2 and \(14.3 ?\) c. What proportion of the population is less than \(10.0 ?\)

Refer to the Real Estate data, which report information on homes sold in the Denver, Colorado, area during the last year. a. The mean selling price (in \(\$$ thousands) of the homes was computed earlier to be \)\$ 221.10,\( with a standard deviation of \)\$ 47.11 .\( Use the normal distribution to estimate the percent of homes selling for more than \)\$ 280.0 .$ Compare this to the actual results. Does the normal distribution yield a good approximation of the actual results? b. The mean distance from the center of the city is 14.629 miles with a standard deviation of 4.874 miles. Use the normal distribution to estimate the number of homes 18 or more miles but less than 22 miles from the center of the city. Compare this to the actual results. Does the normal distribution yield a good approximation of the actual results?

A uniform distribution is defined over the interval from 2 to \(5 .\) a. What are the values for \(a\) and \(b\) ? b. What is the mean of this uniform distribution? c. What is the standard deviation? d. Show that the total area is 1.00 . e. Find the probability of a value more than 2.6 f. Find the probability of a value between 2.9 and 3.7 .

The annual commissions earned by sales representatives of Machine Products, Inc., \(a\) manufacturer of light machinery, follow the normal probability distribution. The mean yearly amount earned is \(\$ 40,000\) and the standard deviation is \(\$ 5,000\). a. What percent of the sales representatives earn more than \(\$ 42,000\) per year? b. What percent of the sales representatives earn between \(\$ 32,000\) and \(\$ 42,000 ?\) c. What percent of the sales representatives earn between \(\$ 32,000\) and \(\$ 35,000 ?\) d. The sales manager wants to award the sales representatives who earn the largest commissions a bonus of \(\$ 1,000\). He can award a bonus to 20 percent of the representatives. What is the cutoff point between those who earn a bonus and those who do not?

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