/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 Miles driven A taxicab company i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Miles driven A taxicab company in New York City analyzed the daily number of miles driven by each of its drivers. It found the average distance was 200 mi with a standard deviation of 30 mi. Assuming a normal distribution, what prediction can we make about the percentage of drivers who will log in either more than 260 mi or less than 170 \(\mathrm{mi} ?\)

Short Answer

Expert verified
Approximately 18.15% of drivers log either more than 260 miles or less than 170 miles.

Step by step solution

01

Understanding the Problem

We need to determine the percentage of drivers who log either more than 260 miles or less than 170 miles, given an average of 200 miles and a standard deviation of 30 miles with a normal distribution.
02

Calculate Z-scores

Convert the mile ages into Z-scores using the formula: \[ Z = \frac{X - \mu}{\sigma} \]where \(X\) is the value, \(\mu\) is the mean (200 miles), and \(\sigma\) is the standard deviation (30 miles). For 260 miles:\[ Z_{260} = \frac{260 - 200}{30} = 2 \]For 170 miles:\[ Z_{170} = \frac{170 - 200}{30} = -1 \]
03

Use the Z-table

Using the Z-table, find the percentage of data that falls below each Z-score. For \(Z = 2\), approximately 97.72% of data falls below. For \(Z = -1\), approximately 15.87% of data falls below.
04

Calculate Percentages

Determine the percentage of drivers above 260 miles:\[ 100ackslash%-97.72ackslash% = 2.28ackslash% \]Determine the percentage of drivers below 170 miles, which is 15.87%.
05

Combine Results

Add the percentages together to find the percentage of drivers outside the 170-260 mile range:\[ 2.28ackslash% + 15.87ackslash% = 18.15ackslash% \]
06

Conclusion

The prediction is that approximately 18.15% of drivers log either more than 260 miles or less than 170 miles based on the normal distribution.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Z-scores
Z-scores are an essential concept in statistics, especially when dealing with normal distributions. They help us understand how far a particular data point is from the mean by telling us how many standard deviations away it is. This is important because it allows us to compare data points in different contexts or datasets.

The formula to calculate a Z-score is\[ Z = \frac{X - \mu}{\sigma} \]where
  • \(X\) is the data point in question,
  • \(\mu\) is the mean of the data set, and
  • \(\sigma\) is the standard deviation.
For example, in the exercise where we calculated the Z-scores for 260 miles and 170 miles, we used the given mean of 200 miles and standard deviation of 30 miles. This helped us determine whether these mileages were significantly different from the average distance driven.
Standard Deviation
Standard deviation is a crucial concept in statistics that measures the amount of variation or dispersion in a set of data. In simpler terms, it tells us how spread out the numbers in a data set are from the mean.

When data points are closer to the mean, the standard deviation is small. Conversely, when data points are more spread out from the mean, the standard deviation is larger.

In the context of our exercise, a standard deviation of 30 miles indicates the typical amount of variation from the average distance (200 miles) that drivers in New York City might travel each day. Understanding standard deviation helps us predict how data points (such as miles driven) deviate from the average.
Mean
The mean, often referred to as the average, is a measure of central tendency that sums up a set of values and divides by the number of values. It's a useful value to understand the general "center" of a dataset.

In our exercise's case, the mean number of miles driven by a taxi driver per day was 200 miles. This mean is a baseline from which we calculate Z-scores and understand the data's variance via standard deviation.

Understanding the mean provides a context for analyzing a normal distribution, allowing predictions like calculating the percentage of drivers who log more than 260 miles or fewer than 170 miles. It is a foundational element in understanding any statistical analysis or prediction based on the normal distribution.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$\int_{1}^{2} \frac{d x}{x \ln x}$$

Customer service time The The mean waiting time to get served after walking into a bakery is 30 seconds. Assume that an exponential density function describes the waiting times. $$ \begin{array}{l}{\text { a. What is the probability a customer waits } 15 \text { seconds or less? }} \\ {\text { b. What is the probability a customer waits longer than one }} \\ {\text { minute? }} \\ {\text { c. What is the probability a customer waits exactly } 5 \text { minutes? }} \\ {\text { d. If } 200 \text { customers come to the bakery in a day, how many are }} \\\ {\text { likely to be served within three minutes? }}\end{array} $$

Blood pressure Diastolic blood pressure in adults is normally distributed with \(\mu=80 \mathrm{mm} \mathrm{Hg}\) and \(\sigma=12 \mathrm{mm} \mathrm{Hg} .\) In a random sample of 300 adults, how many would be expected to have a diastolic blood pressure below 70 \(\mathrm{mm}\) ?

Use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$\int_{4}^{\infty} \frac{d x}{\sqrt{x}-1}$$

Show that if \(f(x)\) is integrable on every interval of real numbers and \(a\) and \(b\) are real numbers with \(a < b\) , then \(\begin{equation} \begin{array}{l}{\text { a. } \int_{-\infty}^{a} f(x) d x \text { and } \int_{a}^{\infty} f(x) d x \text { both converge if and only if }} \\ \quad {\int_{-\infty}^{b} f(x) d x \text { and } \int_{b}^{\infty} f(x) d x \text { both converge. }} \\ {\text { b. } \int_{-\infty}^{a} f(x) d x+\int_{a}^{\infty} f(x) d x=\int_{-\infty}^{b} f(x) d x+\int_{b}^{\infty} f(x) d x} \\ {\quad \text { when the integrals involved converge. }}\end{array} \end{equation}\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.