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The number of vehicles leaving a turnpike at a certain exit during a particular time period has approximately a normal distribution 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
The probability that at least 650 cars leaving the turnpike is approximately 0.0228 or 2.28%, strictly between 400 and 550 cars are 0.4374 or 43.74%, and between 400 and 550 cars (inclusive) are also approximately 0.4374 or 43.74%.

Step by step solution

01

Identifying Parameters

This exercise is related to a standard normal curve which is often referred to as a normal distribution. The mean (\(\mu\)) is 500 and the standard deviation (\(\sigma\)) is 75.
02

Finding Z-score for 650

The Z-score for a value x is given by the equation \[ Z = (X - \mu)/\sigma \] Where x is the value, \(\mu\) is the mean and \(\sigma\) is the standard deviation. Calculate Z-score for x=650: \[ Z = (650 - 500)/75 \]
03

Looking up Probability P(Z

By consulting a Z-table or using statistical software, we can find the probability \$P(Z<Z\_value\$).
04

Finding P(Z>2)

Since we are looking for the chance that there are at least 650 cars, which corresponds to \(Z > 2\), we need to find \(P(Z > 2)\). The Z-table gives the values for \(P(Z < z)\), so to find \(P(Z > z)\) we subtract the tabled value from 1. Therefore, \(P(Z > 2) = 1 - P(Z < 2)\). Check the table for \(P(Z < 2)\) and perform the calculation.
05

Finding Z-score for 400 and Z-score for 550

Calculate Z-score for x=400 and x=550 to find boundaries for the next steps. \[ Z_1 = (400 - 500)/75 \] \[ Z_2 = (550 - 500)/75 \]
06

Looking up Probability P(Z1

By consulting a Z-table or using statistical software, we can find the probability \(P(Z1<Z<Z2\).
07

Finding P(-1.33

Since we are given that the number of cars is strictly between 400 and 550, which corresponds to \(-1.33<Z<0.67\), and also inclusively between these numbers, we need to find \(P(-1.33<Z<0.67)\) and \(P(-1.33<=Z<=0.67)\). Check the table for \(P(Z < 0.67)\) and \(P(Z < -1.33)\) and perform the calculation.

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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 crucial when dealing with normal distributions. In essence, a Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. If you have a Z-score of 0, it indicates that the data point's score is identical to the mean score.

To calculate the Z-score of a particular data point, you use the formula: \[ Z = \frac{(X - \mu)}{\sigma} \] where \(X\) is the value you're interested in, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. For instance, in the textbook example, to find the Z-score for 650 cars exiting the turnpike, you would subtract the mean (500) from 650 and then divide by the standard deviation (75), resulting in a Z-score that indicates how many standard deviations 650 cars is from the mean.
Standard Normal Curve
The standard normal curve, or the bell curve, is a graphical representation of a normal probability distribution. It is symmetrical, with its peak corresponding to the mean value. The properties of the curve are such that approximately 68% of the data falls within one standard deviation from the mean, 95% within two, and 99.7% within three standard deviations.

When working with standard normal distribution, one typically uses a Z-score to navigate the curve. This process standardizes any normal distribution, with mean 0 and standard deviation of 1, and makes it possible to compare different data sets by bringing them onto a common scale. In educational practices, this concept is often underscored with diagrams depicting the curve and shading the relevant sections based on the Z-scores.
Probability Distribution Tables
Probability distribution tables, commonly known as Z-tables, are tools that allow us to find the probability of a Z-score occurring within a standard normal distribution. These tables show the cumulative probability associated with a Z-score; for instance, they indicate the probability that a random variable will be less than a particular Z-value.

To use a Z-table, you first calculate the Z-score of the data point in question. You can then refer to the table to find the corresponding probability. It is important to note that Z-tables provide the probability that a value is less than the Z-score (\(P(Z < z)\)). Therefore, if you need the probability of a value being more than the Z-score (\(P(Z > z)\)), you would subtract the table probability from 1. Teachers often encourage students to have a copy of the Z-table when studying probability as it's a practical tool for solving a variety of statistical problems.
Statistical Parameters
Statistical parameters are the core numbers obtained from a data set—the most common of which are the mean, median, mode, variance, and standard deviation. They provide key insights into the characteristics of the data.

The mean tells us about the central point of the data. The standard deviation, which appears in our Z-score formula, indicates the amount of variability or dispersion in a set of values. A low standard deviation suggests the data points are close to the mean, while a high standard deviation indicates the data points are spread out over a broader range of values. Diving into the realm of statistics, the mastery of these parameters lays the groundwork for deeper and more complex analysis, such as hypothesis testing and regression analysis.

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Most popular questions from this chapter

The longest "run" of \(S\) 's in the sequence SSFSSSSFFS has length 4 , corresponding to the \(S\) 's on the fourth, fifth, sixth, and seventh trials. Consider a binomial experiment with \(n=4\), and let \(y\) be the length (number of trials) in the longest run of \(S\) 's. a. When \(\pi=.5\), the 16 possible outcomes are equally likely. Determine the probability distribution of \(y\) in this case (first list all outcomes and the \(y\) value for each one). Then calculate \(\mu_{y}\) b. Repeat Part (a) for the case \(\pi=.6\). c. Let \(z\) denote the longest run of either \(S\) 's or \(F\) 's. Determine the probability distribution of \(z\) when \(\pi=.5\).

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A multiple-choice exam consists of 50 questions. Each question has five choices, of which only one is correct. Suppose that the total score on the exam is computed as $$ y=x_{1}-\frac{1}{4} x_{2} $$ where \(x_{1}=\) number of correct responses and \(x_{2}=\) number of incorrect responses. (Calculating a total score by subtracting a term based on the number of incorrect responses is known as a correction for guessing and is designed to discourage test takers from choosing answers at random.) a. It can be shown that if a totally unprepared student answers all 50 questions by just selecting one of the five answers at random, then \(\mu_{x_{1}}=10\) and \(\mu_{x_{2}}=40\). What is the mean value of the total score, \(y\) ? Does this surprise you? Explain. b. Explain why it is unreasonable to use the formulas given in this section to compute the variance or standard deviation of \(y\).

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Suppose that the \(\mathrm{pH}\) of soil samples taken from a certain geographic region is normally distributed with a mean \(\mathrm{pH}\) of \(6.00\) and a standard deviation of \(0.10 .\) If the \(\mathrm{pH}\) of a randomly selected soil sample from this region is determined, answer the following questions about it: a. What is the probability that the resulting \(\mathrm{pH}\) is between \(5.90\) and \(6.15 ?\) b. What is the probability that the resulting \(\mathrm{pH}\) exceeds \(6.10 ?\) c. What is the probability that the resulting \(\mathrm{pH}\) is at most 5.95? d. What value will be exceeded by only \(5 \%\) of all such pH values?

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