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The distribution of passenger vehicle speeds traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour. \(^{56}\) (a) What percent of passenger vehicles travel slower than 80 miles/hour? (b) What percent of passenger vehicles travel between 60 and 80 miles/hour? (c) How fast do the fastest \(5 \%\) of passenger vehicles travel? (d) The speed limit on this stretch of the I-5 is 70 miles/hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the I-5.

Short Answer

Expert verified
(a) 93.88%, (b) 93.44%, (c) 80.46 mph, (d) 70.45%

Step by step solution

01

Analyze the Given Information

We have a nearly normal distribution of vehicle speeds with a mean, \(\mu = 72.6\) mph, and a standard deviation, \(\sigma = 4.78\) mph. We need to find probabilities and values based on this distribution.
02

Calculate Z-Score for 80 mph (Part a)

The Z-score is calculated by the formula \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the value of interest. For 80 mph, \( Z = \frac{80 - 72.6}{4.78} \approx 1.54 \).
03

Determine Percentage for Z-Score (Part a)

Using Z-tables or a standard normal distribution calculator, find the percentage of vehicles traveling slower than 80 mph. A Z-score of 1.54 corresponds to approximately 93.88%.
04

Calculate Z-Score for 60 mph (Part b)

For 60 mph, compute \( Z = \frac{60 - 72.6}{4.78} \approx -2.63 \).
05

Find Percent between 60 and 80 mph (Part b)

Find the cumulative probability for \( Z = -2.63 \), which is approximately 0.44%, and subtract it from the cumulative probability at 80 mph, which is approximately 93.88%. Thus, approximately 93.44% of vehicles travel between 60 mph and 80 mph.
06

Find the Z-Score for Fastest 5% (Part c)

The top 5% vehicle speeds correspond to a Z-score where 95% are below. This Z-score is approximately 1.645 from standard normal distribution tables.
07

Calculate Speed for Z = 1.645 (Part c)

Using the Z-score formula \( X = \mu + Z\sigma \), compute \( X \approx 72.6 + 1.645 \times 4.78 \approx 80.46 \) mph. The fastest 5% of vehicles travel faster than 80.46 mph.
08

Find Z-Score for 70 mph (Part d)

Calculate the Z-score for 70 mph, \( Z = \frac{70 - 72.6}{4.78} \approx -0.54 \).
09

Determine Percentage Above 70 mph (Part d)

Use Z-tables or calculators to find the cumulative probability for \( Z = -0.54 \), approximately 29.55%. Therefore, 100% - 29.55% = 70.45% of vehicles travel above 70 mph.

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Key Concepts

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

Z-score calculation
In statistics, a Z-score indicates how many standard deviations an element is from the mean. It's a way to understand the data point's position relative to the overall data distribution.
You calculate a Z-score using the formula:
  • \( Z = \frac{X - \mu}{\sigma} \)
  • Where \( X \) is the value of interest
  • \( \mu \) is the mean
  • \( \sigma \) is the standard deviation
For example, to find out how an 80 mph speed compares to a distribution with a mean speed of 72.6 mph and a standard deviation of 4.78 mph, you use:
  • \( Z = \frac{80 - 72.6}{4.78} \approx 1.54 \)
This result tells us that 80 mph is 1.54 standard deviations above the mean, helping us infer how common or rare it is in this context.
Understanding Z-scores is crucial for comparing data from different normal distributions and for finding how extreme a data point is compared to the rest of the dataset.
Standard Deviation
Standard deviation is a statistical measure that describes the amount of variation or dispersion in a set of values. It provides insights into the consistency or spread of the data points in relation to the mean. A smaller standard deviation indicates that the data points are closer to the mean, reflecting less variability. In contrast, a larger standard deviation suggests more spread out data.
To compute the standard deviation, one often uses the formula:
  • Calculate the mean (average) of the dataset.
  • Subtract the mean from each data point and square the result.
  • Calculate the average of those squared differences.
  • Take the square root of that average.
In our vehicle speed example, standard deviation (\( \sigma \)) is 4.78 mph. It reflects how widely or tightly speeds vary around the mean speed of 72.6 mph on the freeway. This measure helps identify how speeds deviate, allowing us to gauge speed consistency or variability in our data.
Cumulative Probability
Cumulative probability is the probability that a random variable takes on a value less than or equal to a specific value. In the context of normal distribution, this refers to the area under the curve to the left of a given point.
To determine cumulative probability, Z-tables or a standard normal distribution calculator can be used. They provide areas or probabilities associated with different Z-scores.
For example, if a Z-score is 1.54 (from a previous calculation for a speed of 80 mph), the cumulative probability is approximately 93.88%. This implies that about 93.88% of vehicles travel at speeds less than or equal to 80 mph. These probabilities are critical in decision-making and predicting outcomes based on the data's behavior.
Such probabilities enable us to interpret data trends and to better understand the proportion of data points within specific ranges in a normal distribution.

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Most popular questions from this chapter

In 2013 , the Pew Research Foundation reported that " \(45 \%\) of U.S. adults report that they live with one or more chronic conditions", and the standard error for this estimate is \(1.2 \%\). Identify each of the following statements as true or false. Provide an explanation to justify each of your answers. (a) We can say with certainty that the confidence interval from Exerise 2.37 contains the true percentage of U.S. adults who suffer from a chronic illness. (b) If we repeated this study 1,000 times and constructed a \(95 \%\) confidence interval for each study, then approximately 950 of those confidence intervals would contain the true fraction of U.S. adults who suffer from chronic illnesses. (c) The poll provides statistically significant evidence (at the \(\alpha=0.05\) level) that the percentage of U.S. adults who suffer from chronic illnesses is below \(50 \%\). (d) Since the standard error is \(1.2 \%\), only \(1.2 \%\) of people in the study communicated uncertainty about their answer.

Assisted Reproductive Technology (ART) is a collection of techniques that help facilitate pregnancy (e.g. in vitro fertilization). A 2008 report by the Centers for Disease Control and Prevention estimated that ART has been successful in leading to a live birth in \(31 \%\) of cases \(^{55}\). A new fertility clinic claims that their success rate is higher than average. A random sample of 30 of their patients yielded a success rate of \(40 \%\). A consumer watchdog group would like to determine if this provides strong evidence to support the company's claim. (a) Write the hypotheses to test if the success rate for ART at this clinic is significantly higher than the success rate reported by the CDC. (b) Describe a setup for a simulation that would be appropriate in this situation and how the p-value can be calculated using the simulation results. (c) Below is a histogram showing the distribution of \(\hat{p}_{s i m}\) in 10,000 simulations under the null hypothesis. Estimate the p-value using the plot and use it to evaluate the hypotheses. (d) After performing this analysis, the consumer group releases the following news headline: "Infertility clinic falsely advertises better success rates". Comment on the appropriateness of this statement.

Find the standard deviation of the distribution in the following situations. (a) MENSA is an organization whose members have IQs in the top \(2 \%\) of the population. IQs are normally distributed with mean 100 , and the minimum IQ score required for admission to MENSA is 132 . (b) Cholesterol levels for women aged 20 to 34 follow an approximately normal distribution with mean 185 milligrams per deciliter \((\mathrm{mg} / \mathrm{dl})\). Women with cholesterol levels above \(220 \mathrm{mg} / \mathrm{dl}\) are considered to have high cholesterol and about \(18.5 \%\) of women fall into this category.

A college senior who took the Graduate Record Examination exam scored 620 on the Verbal Reasoning section and 670 on the Quantitative Reasoning section. The mean score for Verbal Reasoning section was 462 with a standard deviation of \(119,\) and the mean score for the Quantitative Reasoning was 584 with a standard deviation of 151. Suppose that both distributions are nearly normal. (a) Write down the short-hand for these two normal distributions. (b) What is her \(\mathrm{Z}\) score on the Verbal Reasoning section? On the Quantitative Reasoning section? Draw a standard normal distribution curve and mark these two Z scores. (c) What do these \(Z\) scores tell you? (d) Relative to others, which section did she do better on? (e) Find her percentile scores for the two exams. (f) What percent of the test takers did better than her on the Verbal Reasoning section? On the Quantitative Reasoning section? (g) Explain why simply comparing her raw scores from the two sections would lead to the incorrect conclusion that she did better on the Quantitative Reasoning section. (h) If the distributions of the scores on these exams are not nearly normal, would your answers to parts (b) - (f) change? Explain your reasoning.

What percent of a standard normal distribution \(N(\mu=0, \sigma=\) 1) is found in each region? Be sure to draw a graph. (a) \(Z>-1.13\) (b) \(Z<0.18\) (c) \(Z>8\) (d) \(|Z|<0.5\)

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