/*! 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 58 Suppose that fuel efficiency (mi... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that fuel efficiency (miles per gallon, mpg) for a particular car model under specified conditions is normally distributed with a mean value of \(30.0 \mathrm{mpg}\) and a standard deviation of \(1.2 \mathrm{mpg}\). a. What is the probability that the fuel efficiency for a randomly selected car of this model is between 29 and \(31 \mathrm{mpg}\) ? b. Would it surprise you to find that the efficiency of a randomly selected car of this model is less than \(25 \mathrm{mpg} ?\) c. If three cars of this model are randomly selected, what is the probability that each of the three have efficiencies exceeding \(32 \mathrm{mpg}\) ? d. Find a number \(x^{*}\) such that \(95 \%\) of all cars of this model have efficiencies exceeding \(x^{*}\) (i.e., \(P\left(x>x^{*}\right)=0.95\) ).

Short Answer

Expert verified
The probability that the fuel efficiency for a randomly selected car of this model is between 29 and 31 mpg is 0.5954 or 59.54%. It would be very surprising to find that the efficiency of a randomly selected car of this model is less than 25 mpg. The probability of each of the three cars having efficiency exceeding 32 mpg is approximately 0.0109% or \(1.09 \times 10^{-4}\). 95% of all cars of this model have efficiencies exceeding 28.03 mpg.

Step by step solution

01

Calculate Z-scores

To find the probability that the fuel efficiency is between 29 and 31 mpg, first we need to convert the given fuel efficiencies into Z-scores. The formula to calculate the Z-score is: \(Z = \frac{x - \mu}{\sigma}\) Where \(x\) is the fuel efficiency (in mpg), \(\mu\) is the mean fuel efficiency (30.0 mpg), and \(\sigma\) is the standard deviation (1.2 mpg). Calculate the Z-scores for 29 and 31 mpg: \(Z_{29} = \frac{29 - 30}{1.2} = -0.8333\) \(Z_{31} = \frac{31 - 30}{1.2} = 0.8333\)
02

Find probabilities from Z-table

Now, we have to look up the Z-scores in the Z-table to find the probabilities. For \(Z_{29} = -0.8333\), we find the probability \(P(Z < -0.8333) = 0.2023\). For \(Z_{31} = 0.8333\), we find the probability \(P(Z < 0.8333) = 0.7977\).
03

Calculate the probability between 29 and 31 mpg

To find the probability between 29 and 31 mpg, we need to subtract the smaller probability from the larger probability: \(P(29 < x < 31) = P(Z_{31}) - P(Z_{29}) = 0.7977 - 0.2023 = 0.5954\) So, the probability that the fuel efficiency for a randomly selected car of this model is between 29 and 31 mpg is 0.5954 or 59.54%. b. Probability of efficiency less than 25 mpg
04

Calculate Z-score

We will do the same steps as in the previous problem, but for 25 mpg: \(Z_{25} = \frac{25 - 30}{1.2} = -4.1667\)
05

Find probability from Z-table

However, in a Z-table, the lowest Z-score is about -3.5, which corresponds to a probability of almost 0. If a Z-score is beyond -3.5, we can assume the probability is very close to 0. So, \(P(Z < -4.1667) \approx 0\). It would be very surprising to find that the efficiency of a randomly selected car of this model is less than 25 mpg. c. Probability of three cars having efficiency exceeding 32 mpg
06

Calculate Z-score

Calculate Z-score for 32 mpg: \(Z_{32} = \frac{32 - 30}{1.2} = 1.6667\)
07

Find probability from Z-table

Find probability in the Z-table: \(P(Z > 1.6667) = 1 - P(Z < 1.6667) = 1 - 0.9522 = 0.0478\)
08

Calculate probability for three cars

Since we're looking for the probability that all three cars have efficiencies exceeding 32 mpg, and each car's efficiency is independent, we just need to multiply the probabilities of each car having efficiency above 32 mpg: \(P(\text{3 cars} > 32 \text{mpg}) = P^3(Z > 1.6667) = (0.0478)^3 = 0.000109\) So, the probability of each of the three cars having efficiency exceeding 32 mpg is approximately 0.0109% or \(1.09 \times 10^{-4}\). d. Find 95% efficiency threshold
09

Find Z-score

For a 95% threshold, we want to find the Z-score such that the area to the right of the Z-score is 0.95: \(P(Z > Z^{*}) = 0.95 \Rightarrow P(Z < Z^{*}) = 0.05\) Looking at the Z-table, we find the Z-score that corresponds to a probability of 0.05 is approximately -1.645.
10

Calculate fuel efficiency value

Now, we will use the Z-score formula to find the fuel efficiency value \(x^{*}\) corresponding to this Z-score: -1.645 = \(\frac{x^{*} - 30}{1.2}\) Solving for \(x^{*}\), we get: \(x^{*} = -1.645 \times 1.2 + 30 \approx 28.03\) So, 95% of all cars of this model have efficiencies exceeding 28.03 mpg.

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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 valuable tool when working with normal distributions. A Z-score represents the number of standard deviations a data point, such as a fuel efficiency measurement, is from the mean of the distribution. It's calculated using the formula: \[Z = \frac{x - \mu}{\sigma}\] where \(x\) is the value we want to convert (in this case, miles per gallon), \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

For our exercise, if you want to find where 29 mpg stands in respect to a car's typical performance, you plug in 29 for \(x\), 30 for \(\mu\), and 1.2 for \(\sigma\). The Z-score gives you an easy way to check how unusual a specific measurement is by showing it on a standard normal distribution curve. Lower or higher Z-scores indicate more unusual data points.
Probability Calculations
Probability calculations using Z-scores help us find out how likely (or unlikely) it is for a particular event to occur. In the exercise, we sought probabilities for certain fuel efficiencies:
- The probability of fuel efficiency between two values (e.g., 29 and 31 mpg).
- The rarity of exceptionally low values, like less than 25 mpg.

After computing the Z-scores for these values, you consult a Z-table, a tool that tells you how much of the data falls to the left of a given Z-score in a standard normal distribution. To find the probability that a value is between two points, we simply subtract the smaller probability from the larger one. This provides us with insights into the frequency of those readings, which is particularly useful in understanding everyday likelihoods versus rare readings.
Statistical Thresholds
Statistical thresholds are boundaries in data, computed to understand the geventy of extreme or typical performances. One common threshold is the 95th percentile. In this exercise, we wanted to find a threshold \(x^{*}\), such that 95% of the car model's fuel efficiencies exceed this value.

To find this, we need to identify the corresponding Z-score first. When a Z-score has 95% of the distribution to its right, it means only 5% is to its left. This particular Z-score is around -1.645. By using the Z-score formula backwards, we can solve for \(x^{*}\). \[x^{*} = -1.645 \times 1.2 + 30 \approx 28.03\]
Such thresholds are valuable in quality control, checking if products meet a standard or how infrequently they fail—a necessity in many industries for regulation compliance.

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Most popular questions from this chapter

Studies have found that women diagnosed with cancer in one breast also sometimes have cancer in the other breast that was not initially detected by mammogram or physical examination ("MRI Evaluation of the Contralateral Breast in Women with Recently Diagnosed Breast Cancer," The New England Journal of Medicine [2007]: 1295-1303). To determine if magnetic resonance imaging (MRI) could detect missed tumors in the other breast, 969 women diagnosed with cancer in one breast had an MRI exam. The MRI detected tumors in the other breast in 30 of these women. a. Use \(p=\frac{30}{969}=0.031\) as an estimate of the probability that a woman diagnosed with cancer in one breast has an undetected tumor in the other breast. Consider a random sample of 500 women diagnosed with cancer in one breast. Explain why it is reasonable to think that the random variable \(x=\) number in the sample who have an undetected tumor in the other breast has a binomial distribution with \(n=500\) and \(p=0.031\). b. Is it reasonable to use the normal distribution to approximate probabilities for the random variable \(x\) defined in Part (a)? Explain why or why not. c. Approximate the following probabilities: i. \(\quad P(x<10)\) ii. \(\quad P(10 \leq x \leq 25)\) iii. \(P(x>20)\) d. For each of the probabilities calculated in Part (c), write a sentence interpreting the probability.

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