91Ó°ÊÓ

Based on tests of the Chevrolet Cobalt, engineers have found that the miles per gallon in highway driving are normally distributed, with a mean of 32 miles per gallon and a standard deviation 3.5 miles per gallon. (a) What is the probability that a randomly selected Cobalt gets more than 34 miles per gallon? (b) Suppose that 10 Cobalts are randomly selected and the miles per gallon for each car are recorded. What is the probability that the mean miles per gallon exceed 34 miles per gallon? (c) Suppose that 20 Cobalts are randomly selected and the miles per gallon for each car are recorded. What is the probability that the mean miles per gallon exceed 34 miles per gallon? Would this result be unusual?

Step by step solution

01

- Understand the Problem

We need to find the probability that a randomly selected Chevrolet Cobalt gets more than 34 miles per gallon (mpg) given that the mpg is normally distributed with a mean (µ) of 32 and a standard deviation (σ) of 3.5.

02

- Convert the Value to a Z-score (Part a)

First, we convert 34 mpg to a Z-score using the formula: \[ Z = \frac{{X - µ}}{{σ}} \] Here, \( X = 34 \), \( µ = 32 \), and \( σ = 3.5 \). Substitute these numbers in: \[ Z = \frac{{34 - 32}}{{3.5}} = \frac{2}{3.5} = 0.5714 \]

03

- Find the Probability (Part a)

Using the Z-score table or a standard normal distribution calculator, find the probability for \( P(Z > 0.5714) \). The table or calculator tells us that \( P(Z < 0.5714) = 0.7166 \). Therefore, \( P(Z > 0.5714) \) is: \[ P(Z > 0.5714) = 1 - P(Z < 0.5714) = 1 - 0.7166 = 0.2834 \]

04

- Mean and Standard Deviation of the Sampling Distribution (Part b)

When 10 Cobalts are selected, the sampling distribution of the mean mpg has a mean (µ) of 32 and a standard deviation (σ) given by: \[ σ_{\bar{X}} = \frac{σ}{\sqrt{n}} \] where \( n = 10 \). Calculate the standard deviation: \[ σ_{\bar{X}} = \frac{3.5}{\sqrt{10}} = \frac{3.5}{3.1623} = 1.1068 \]

05

- Convert the Mean to a Z-score (Part b)

Convert the mean of 34 mpg to a Z-score using: \[ Z = \frac{\bar{X} - μ}{σ_{\bar{X}}} \] Here, \( \bar{X} = 34 \), \( μ = 32 \), and \( σ_{\bar{X}} = 1.1068 \). Substitute the values: \[ Z = \frac{34 - 32}{1.1068} = \frac{2}{1.1068} = 1.807 \]

06

- Find the Probability (Part b)

Using the Z-score table or calculator, find the probability for \( P(Z > 1.807) \). The table tells us that \( P(Z < 1.807) = 0.9649 \). Therefore, \( P(Z > 1.807) \) is: \[ P(Z > 1.807) = 1 - 0.9649 = 0.0351 \]

07

- Repeat Calculations for 20 Cars (Part c)

When 20 Cobalts are selected, use similar steps. The standard deviation for the sample mean is: \[ σ_{\bar{X}} = \frac{3.5}{\sqrt{20}} = \frac{3.5}{4.472} = 0.7826 \] Convert 34 mpg to Z-score: \[ Z = \frac{34 - 32}{0.7826} = \frac{2}{0.7826} = 2.556 \]

08

- Find the Probability (Part c)

Using the Z-score table or calculator, find the probability for \( P(Z > 2.556) \). The table tells us that \( P(Z < 2.556) = 0.9947 \). Therefore, \( P(Z > 2.556) \) is: \[ P(Z > 2.556) = 1 - 0.9947 = 0.0053 \] This probability is very low and would be considered unusual.

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.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. Think of it as a way to estimate how spread out the values in a data set are around the mean.
In the context of the exercise, the standard deviation (σ) is given as 3.5 miles per gallon (mpg). This means the fuel efficiencies of the Chevrolet Cobalts will, on average, deviate by about 3.5 mpg from the mean of 32 mpg.
Understanding standard deviation helps us to describe how varied our data points are. A small standard deviation means the data points are close to the mean, whereas a large standard deviation indicates the data points are spread out over a wider range.

Z-Score

A Z-score, also known as a standard score, measures how many standard deviations an element is from the mean. It's a way of standardizing different values onto a common scale. The formula for calculating a Z-score is:
\( Z = \frac{X - \mu}{\sigma} \)
In the exercise, we calculated the Z-score for 34 miles per gallon as follows: \( Z = \frac{34 - 32}{3.5} = 0.5714 \).
This Z-score tells us that 34 mpg is 0.5714 standard deviations above the mean. Understanding Z-scores is crucial for probability calculations, especially when dealing with the normal distribution.

Sampling Distribution

A sampling distribution refers to the probability distribution of a given statistic based on a random sample. It's important when comparing sample means. The mean of a sampling distribution is the same as the population mean, but the standard deviation (standard error) changes depending on the sample size (n). The formula for the standard error of the mean is:
\( \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} \)
In the exercise, for 10 cars, the standard error was calculated as: \( \sigma_{\bar{X}} = \frac{3.5}{\sqrt{10}} = 1.1068 \).
Knowing the sampling distribution helps in making inferences about the population mean from sample means.

Probability Calculation

Probability calculation in the context of the normal distribution involves finding areas under the curve. After converting values to Z-scores, we use Z-tables or calculators to find the probability.
For instance, to find the probability that a car has more than 34 mpg, we use the Z-score of 0.5714. The probability corresponding to this Z-score is 0.7166, so the probability of getting more than 34 mpg is \( 1 - 0.7166 = 0.2834 \).
Moreover, when dealing with sample means, similar steps follow, but using the standard error as seen in parts (b) and (c) of the exercise. The process still involves Z-scores and using tables or calculators to find the probability above a certain value.