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The number of vehicles leaving a highway at a certain exit during a particular time period has a distribution that is approximately normal with mean value 500 and standard deviation \(75 .\) What is the probability that the number of cars exiting during this period is a. At least \(650 ?\) b. Strictly between 400 and \(550 ?\) (Strictly means that the values 400 and 550 are not included.) c. Between 400 and 550 (inclusive)?

Short Answer

Expert verified
a. The probability that the number of vehicles exiting during the period is at least 650 is 0.0228. b. The probability that the number of vehicles exiting during the period is strictly between 400 and 550 is 0.6568. c. The probability that the number of vehicles exiting during the period is between 400 and 550 (inclusive) is 0.6568.

Step by step solution

01

Identify the mean and standard deviation

The problem states that the distribution is approximately normal with a mean (μ) of 500 vehicles and a standard deviation (σ) of 75.
02

Calculate the z-scores for each part of the problem

We will calculate the z-score for each part of the problem: a. At least 650 cars b. Strictly between 400 and 550 c. Between 400 and 550 (inclusive) Now, recall the formula for calculating the z-score: \( z = \frac{x - μ}{σ} \) For part (a): \( z = \frac{650 - 500}{75} \) For part (b): \( z_1 = \frac{400 - 500}{75} \) and \( z_2 = \frac{550 - 500}{75} \) For part (c): Since the values are inclusive, we don't need to change the z-scores from part (b): \( z_1 = \frac{400 - 500}{75} \) and \( z_2 = \frac{550 - 500}{75} \)
03

Calculate the z-scores

Now, we will calculate the z-scores for each part: For part (a): \( z = \frac{650 - 500}{75} = 2 \) For part (b): \( z_1 = \frac{400 - 500}{75} = -\frac{4}{3} \approx -1.33 \) \( z_2 = \frac{550 - 500}{75} = \frac{2}{3} \approx 0.67 \) For part (c): Since the values are inclusive, we don't need to change the z-scores from part (b): \( z_1 = -1.33 \) and \( z_2 = 0.67 \)
04

Find probabilities using standard normal distribution table

Next, we will consult the standard normal distribution table to find the probabilities corresponding to these z-scores: For part (a): The probability of having a z-score less than 2 is 0.9772. But we want the probability of having a z-score greater than 2, so we subtract the probability from 1: \( P(z > 2) = 1 - 0.9772 = 0.0228 \) For part (b): To find the probability of being strictly between the two z-scores, we subtract the lower probability from the higher probability: \( P(-1.33 < z < 0.67) = P(z < 0.67) - P(z < -1.33) = 0.7486 - 0.0918 = 0.6568 \) For part (c): Since the values are inclusive, we don't need to change the probabilities from part (b): \( P(-1.33 \leq z \leq 0.67) = 0.6568 \)
05

State the probability for each part of the problem

Finally, we present the probabilities for each part of the problem: a. The probability that the number of vehicles exiting during the period is at least 650 is 0.0228. b. The probability that the number of vehicles exiting during the period is strictly between 400 and 550 is 0.6568. c. The probability that the number of vehicles exiting during the period is between 400 and 550 (inclusive) is 0.6568.

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Key Concepts

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

Z-Score Calculation
Understanding the z-score is fundamental for students navigating the field of probability and statistics. A z-score, also known as a standard score, quantifies the number of standard deviations a data point is from the mean of a distribution. In the given exercise, the z-score is used to transform individual vehicle exit counts into a standardized form that allows comparison on a normal distribution scale. The formula for the z-score is given by:
\[ z = \frac{{x - \mu}}{{\sigma}} \]
where
  • \( x \) is the value from the dataset (e.g., the number of cars exiting)
  • \( \mu \) represents the mean of the dataset (e.g., 500 vehicles)
  • \( \sigma \) is the standard deviation of the dataset (e.g., 75 vehicles)
By employing the z-score calculation, students can easily determine how unusual or typical a particular value is within the data set governed by the normal distribution.
Standard Normal Distribution Table
Once a student calculates the z-score, the next phase is to understand what it signifies in terms of probability. This is where the standard normal distribution table, also known as a z-table, comes into play. This table reflects the probability that a standard normal variable will fall between the mean and any number of standard deviations away from the mean (the z-score).
In our textbook example, the z-scores calculated for different scenarios are converted into probabilities using this table. For instance, a positive z-score (such as 2 from part a) allows you to find the area under the curve to the left of that z-score. Conversely, a negative z-score (such as -1.33 from part b) represents an area to the right. The table then provides the cumulative probability for those areas. In problems that deal with ranges (like parts b and c in the exercise), the probabilities of each z-score are subtracted to find the probability that a value falls within that range. Understanding how to read this table is crucial in analyzing and interpreting data under the standard normal distribution.
Probability and Statistics
At its core, the field of probability and statistics is about making sense of data using mathematical tools. Probability is the measure of the likelihood that an event will occur, and it is quantified as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.
Statistics often deals with analyzing a collection of data and drawing conclusions about that data. This sometimes involves normal distribution, which is a continuous probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.
In the exercise provided, calculating the probability of the number of vehicles exiting the highway within a specific range is a typical statistical problem. By identifying that the number of vehicles is normally distributed, we can harness the z-score calculation and the standard normal distribution table to solve for these probabilities, illustrating an intersection of probability and statistics. Mastery of these principles enables students to apply statistical analysis in various real-world situations beyond the textbook, from science experiments to market research.

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