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Chapter 5: Problem 3

Assume the given distributions are approximately normal. A trucking firm determines that its fleet of trucks averages a mean of 12.4 miles per gallon with a standard deviation of 1.2 miles per gallon on cross-country hauls. What is the probability that one of the trucks averages fewer than 10 miles per gallon? (A) \(P(z<-2.4)\) (B) \(P(z<-2)\) (C) \(P(z<10)\) (D) \(P\left(z<\frac{10}{1.2}\right)\) (E) \(P\left(z<\frac{12.4}{1.2}\right)\)

Short Answer

Expert verified
(B) \(P(z < -2)\)

Step by step solution

01

Identify the given values

The mean \(\mu\) is 12.4 miles per gallon and the standard deviation \(\sigma\) is 1.2 miles per gallon. We need to find the probability that a truck averages fewer than 10 miles per gallon.
02

Write the probability statement

We are looking for \(P(X < 10)\), where \(X\) is the number of miles per gallon.
03

Convert the raw score to a z-score

Use the z-score formula: \[ z = \frac{X - \mu}{\sigma} \] Plugging in the values: \[ z = \frac{10 - 12.4}{1.2} = -2 \]
04

Find the corresponding probability

The probability that a truck averages fewer than 10 miles per gallon is given by \( P(z < -2) \).
05

Compare with the given options

From the options provided, (B) \( P(z < -2) \) matches our result.

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Key Concepts

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

Z-Score Calculation
When dealing with normal distributions, one essential skill is calculating the z-score. The z-score helps you understand how far a specific data point is from the mean, in terms of the standard deviation. The formula for z-score calculation is: \[ z = \frac{X - \mu}{\sigma} \]Here,
    * \(X\) is your data point * \(\mu\) is the mean * \(\sigma\) is the standard deviation
In our exercise, we're given the mean (\(\mu = 12.4\)) and the standard deviation (\(\sigma = 1.2\)). We needed to find the z-score when \(X = 10\). Plugging in the values, we got:\[ z = \frac{10 - 12.4}{1.2} = -2 \]This tells us that 10 miles per gallon is 2 standard deviations below the mean of 12.4 miles per gallon.
Standard Deviation
Standard deviation is a measure of how spread out the numbers in a data set are. In a normal distribution, about 68% of the data falls within one standard deviation of the mean, 95% within two, and 99.7% within three. In our problem,
    * The mean (\(\mu\)) is 12.4 miles per gallon * The standard deviation (\(\sigma\)) is 1.2 miles per gallon
Knowing the standard deviation helps us determine the z-score, which in turn is key to finding probabilities. A smaller standard deviation means the data points are close to the mean, while a larger one means they鈥檙e more spread out. So, the standard deviation of 1.2 shows us that truck fuel efficiencies are relatively consistent but can vary by that amount.
Probability
Probability helps us make predictions based on data. When dealing with normal distributions and z-scores, we frequently want to know the probability of a certain outcome. Once we find the z-score, we can use standard normal distribution tables or software to find this probability.In our case, after calculating the z-score as -2, we need \(P(z < -2)\). This value tells us the probability that a truck averages fewer than 10 miles per gallon. From standard normal distribution tables, \(P(z < -2)\) is approximately 0.0228 or 2.28%. This means there's only a 2.28% chance that a truck will perform below 10 miles per gallon.

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Most popular questions from this chapter

Which of the following is an incorrect statement? (A) The sampling distribution of \(\bar{x}\) has mean equal to the population mean \(\mu\) even if the population is not normally distributed. (B) The sampling distribution of \(\bar{x}\) has standard deviation \(\frac{\sigma}{\sqrt{n}}\) even if the population is not normally distributed. (C) The sampling distribution of \(\bar{x}\) is normal if the population has a normal distribution. (D) When \(n\) is large, the sampling distribution of \(\bar{x}\) is approximately normal even if the population is not normally distributed. (E) The larger the value of the sample size \(n\), the closer the standard deviation of the sampling distribution of \(\bar{x}\) is to the standard deviation of the population.

Assume the given distributions are roughly normal. The mean income per household in a certain state is \(\$ 9500\) with a standard deviation of \(\$ 1750 .\) The middle \(95 \%\) of incomes are between what two values? (A) \(9500 \pm 1.645(1750)\) (B) \(9500 \pm 1.96(1750)\) (C) \(9500 \pm 1.645\left(\frac{1750}{2}\right)\) (D) \(9500 \pm \frac{1750}{1.645}\) (E) \(9500 \pm \frac{1750}{1.96}\)

Which of the following is the best reason that the sample maximum is not used as an estimator for the population maximum? (A) The sample maximum is biased. (B) The sampling distribution of the sample maximum is not binomial. (C) The sampling distribution of the sample maximum is not normal. (D) The sampling distribution of the sample maximum has too large a standard deviation. (E) The sample mean plus three sample standard deviations gives the best estimate for the population maximum.

In a school of 25 oo students, the students in an AP Statistics class are planning a random survey of 100 students to estimate the proportion who would rather drop lacrosse than band during this time of severe budget cuts. Their teacher suggests instead to survey 200 students in order to (A) reduce bias. (B) reduce variability. (C) increase bias. (D) increase variability. (E) make possible stratification between lacrosse and band.

Assume the given distributions are roughly normal. A coffee machine can be adjusted to deliver any fixed number of ounces of coffee. If the machine has a standard deviation in delivery equal to 0.4 ounce, what should be the mean setting so that an 8-ounce cup will overflow only \(0.5 \%\) of the time? (A) \(8-0.995(0.4)\) (B) \(8+1.96(0.4)\) (C) \(8-1.96(0.4)\) (D) \(8+2.576(0.4)\) (E) \(8-2.576(0.4)\)
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