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Is lack of sleep causing traffic fatalities? A study conducted under the auspices of the National Highway Traffic Safety Administration found that the average number of fatal crashes caused by drowsy drivers each year was 1550 ( Business Week , January 26,2004 ). Assume the annual number of fatal crashes per year is normally distributed with a standard deviation of 300 . a. What is the probability of fewer than 1000 fatal crashes in a year? b. What is the probability that the number of fatal crashes will be between 1000 and 2000 for a year? c. For a year to be in the upper \(5 \%\) with respect to the number of fatal crashes, how many fatal crashes would have to occur?

Short Answer

Expert verified
a. 0.0336; b. 0.8996; c. At least 2044 crashes.

Step by step solution

01

Identify the given parameters

We are given that the average number of fatal crashes per year is \( \mu = 1550 \) and the standard deviation \( \sigma = 300 \). Additionally, the distribution of the number of fatal crashes is assumed to be normal.
02

Calculate probability for fewer than 1000 crashes

To find the probability of fewer than 1000 fatal crashes, we need to compute the Z-score for 1000 using the formula: \( Z = \frac{X - \mu}{\sigma} \). Substituting the values, \( Z = \frac{1000 - 1550}{300} = -1.8333 \). We then use the standard normal distribution table to find the probability corresponding to this Z-score, which is approximately \( P(Z < -1.8333) = 0.0336 \).
03

Calculate probability for 1000 to 2000 crashes

For the number of fatal crashes between 1000 and 2000, calculate the Z-scores for both limits. For 1000: \( Z_1 = \frac{1000 - 1550}{300} = -1.8333 \), and for 2000: \( Z_2 = \frac{2000 - 1550}{300} = 1.5 \). Using the standard normal distribution table, we find: \( P(Z < 1.5) = 0.9332 \) and \( P(Z < -1.8333) = 0.0336 \). Then, \( P(1000 < X < 2000) = 0.9332 - 0.0336 = 0.8996 \).
04

Determine the upper 5% threshold

To find the threshold for the top 5%, we find the Z-score corresponding to the 95th percentile of the standard normal distribution, which is approximately 1.645. Then, we convert this Z-score to the actual number of fatal crashes using the formula: \( X = \mu + Z \times \sigma \). So, \( X = 1550 + 1.645 \times 300 = 2043.5 \). Therefore, at least 2044 fatal crashes must occur for the year to be in the upper 5%.

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Key Concepts

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

Normal Distribution
The normal distribution is a fundamental concept in probability and statistics. It is often called the bell curve due to its distinctive shape, which is symmetric and peaked at the mean. This distribution is important because many natural phenomena, such as human traits and measurement errors, tend to be normally distributed.
In our exercise, the number of fatal crashes per year was assumed to follow a normal distribution with an average of 1550 crashes and a standard deviation of 300 crashes. This model helps us predict the likelihood of different outcomes. For example, determining how many accidents occur that deviate significantly from the average. Understanding the normal distribution enables us to make sense of the variability in data around a central point.
Some key properties of the normal distribution include:
  • The mean, median, and mode of a normal distribution are all equal.
  • It is fully described by its mean (\(\mu\)) and standard deviation (\(\sigma\)).
  • The total area under the curve of a normal distribution is equal to 1.
Z-score
The Z-score is a statistical measurement that describes a value's relation to the mean of a group of values measured in terms of standard deviations. For any given data point, the Z-score indicates how many standard deviations away it is from the mean.
In our case, calculating the Z-score helps us understand the position of a specific number of fatal crashes within the normal distribution of all crashes. The formula used to calculate the Z-score is:\[ Z = \frac{X - \mu}{\sigma} \]where:
  • \(X\)is the value of interest,
  • \(\mu\)is the mean,
  • \(\sigma\)is the standard deviation.
By obtaining a Z-score, we can then use a standard normal distribution table to find probabilities related to our question. For example, a Z-score of -1.8333 in the exercise means the value is 1.8333 standard deviations below the mean.
Fatal Crashes Statistics
Fatal crash statistics provide crucial insights into the safety of transportation systems. They help agencies like the National Highway Traffic Safety Administration to identify patterns and potentially dangerous conditions, such as those caused by drowsy driving.
By understanding the average occurrence and variations in fatal crashes, we can design better safety measures and policies. For instance, knowing that the average number of fatal crashes due to drowsy driving is 1550 helps gauge the effectiveness of ongoing safety efforts, while also highlighting areas for improvement.
In the exercise, fatal crash statistics were used in combination with probability concepts to make predictions. These predictions can help in resource allocation for traffic enforcement and public safety initiatives. Additionally, statistical analysis of these figures often informs technological innovations and regulatory changes aimed at reducing these tragic events.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells us how much the individual data points differ from the mean of the dataset. A small standard deviation indicates that the data points tend to be close to the mean, while a large standard deviation indicates a wide range of values.
In the context of the exercise, the standard deviation of 300 indicates how much the number of fatal crashes typically deviates from the mean of 1550. This helps in understanding the spread and reliability of the data—the larger the standard deviation, the less reliable a single year's data point might be in reflecting the true average.
Standard deviation is particularly useful when comparing the spread of two or more data sets. It provides a way to state how consistent or inconsistent the data is around the mean, which can be critical in fields such as quality control and risk assessment.

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Most popular questions from this chapter

Comcast Corporation is the largest cable television company, the second largest Internet service provider, and the fourth largest telephone service provider in the United States. Generally known for quality and reliable service, the company periodically experiences unexpected service interruptions. On January \(14,2009,\) such an interruption occurred for the Comcast customers living in southwest Florida. When customers called the Comcast office, a recorded message told them that the company was aware of the service outage and that it was anticipated that service would be restored in two hours. Assume that two hours is the mean time to do the repair and that the repair time has an exponential probability distribution. a. What is the probability that the cable service will be repaired in one hour or less? b. What is the probability that the repair will take between one hour and two hours? c. For a customer who calls the Comcast office at 1: 00 P.M., what is the probability that the cable service will not be repaired by 5: 00 p.M.?

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