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Consider the population of all one-gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a normal distribution with mean \(\mu=5 \mathrm{ml}\) and standard deviation \(\sigma=0.2 \mathrm{ml}\) is a reasonable model for the distribution of the variable \(x=\) amount of red dye in the paint mixture Use the normal distribution to calculate the following probabilities. (Hint: See Example \(6.21 .\) ) a. \(P(x<5.0)\) b. \(P(x<5.4)\) c. \(P(x \leq 5.4)\) d. \(P(4.64.5)\) f. \(P(x>4.0)\)

Short Answer

Expert verified
a. \(P(x<5.0) = 0.5\) b. \(P(x<5.4) \approx 0.9772\) c. \(P(x \leq 5.4) \approx 0.9772\) d. \(P(4.64.5) \approx 0.9938\) f. \(P(x>4.0) \approx 1\)

Step by step solution

01

Understand the given information and normal distribution

We are given that the amount of red dye in the paint mixture, \(x\), follows a normal distribution with a mean (\(\mu\)) of 5 ml and a standard deviation (\(\sigma\)) of 0.2 ml. The normal distribution is represented by the function \(N(\mu, \sigma^2)\), so in this case, the distribution is given by \(N(5, 0.2^2)\).
02

Find the z-score

The z-score corresponds to the number of standard deviations the value \(x\) is away from the mean, given by the formula: \[z = \frac{x-\mu}{\sigma}\]We will use this formula to find the z-scores for the given x values in each part of the exercise.
03

Use the standard normal distribution table to find the probabilities

Once we have found the z-scores, we will use the standard normal distribution table (also known as z-table) to find the probabilities. The z-table shows the probabilities for the standard normal distribution.
04

Calculate the probabilities

Now that we have our z-scores and standard normal distribution table, we can calculate the probabilities for each part of the problem. a. \(P(x<5.0)\) We find the z-score for \(x = 5.0\): \[z = \frac{5.0-5}{0.2} = 0\] Using the z-table, we find the probability corresponding to z = 0: \(P(x<5.0) = 0.5\) b. \(P(x<5.4)\) We find the z-score for \(x = 5.4\): \[z = \frac{5.4-5}{0.2} = 2\] Using the z-table, we find the probability corresponding to z = 2: \(P(x<5.4) \approx 0.9772\) c. \(P(x \leq 5.4)\) In this case, we are looking for the probability for \(x\) being less than or equal to 5.4, which is the same as \(P(x<5.4)\). So, the answer is the same as the previous part: \(P(x \leq 5.4) \approx 0.9772\) d. \(P(4.64.5)\) We find the z-score for \(x = 4.5\): \[z = \frac{4.5-5}{0.2} = -2.5\] Using the z-table, we find the probability corresponding to z = -2.5: \(P(x<4.5) \approx 0.0062\). Since we are looking for the probability of \(x>4.5\), we find the complement: \(P(x>4.5) \approx 1 - 0.0062 = 0.9938\) f. \(P(x>4.0)\) We find the z-score for \(x = 4.0\): \[z = \frac{4.0-5}{0.2} = -5\] Using the z-table, we find the probability corresponding to z = -5: \(P(x<4.0) \approx 0\). Since the value corresponding to z = -5 falls off the z-chart, the probability is extremely low. Since we are looking for the probability of \(x>4.0\), we find the complement: \(P(x>4.0) \approx 1 - 0 = 1\)

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Key Concepts

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

Z-Score Calculation
The z-score is a statistical measure that describes the position of a raw score in terms of its distance from the mean, when measured in standard deviation units. To calculate it, you can use the formula:
\[ z = \frac{x-\mu}{\sigma} \]
where \( z \) is the z-score, \( x \) is the value from the data set, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. In the context of the exercise, finding the z-score enables us to transform any given value from the original distribution into a corresponding value on the standard normal distribution, thus simplifying the process of finding probabilities.

Why Calculate a Z-Score?

Understanding the z-score is vital as it allows comparison between different datasets by normalizing the scores. For instance, if you want to compare test scores between two classes with different average performances and spread of scores, the z-score provides a way to determine how a score from one class compares to the average score of the other class.In our paint mixture example, calculating the z-score for a particular amount of red dye helps us understand how unusual (or usual) that amount is in relation to the typical amount expected, as defined by the mean.
Standard Normal Distribution
The standard normal distribution is a specific normal distribution with a mean of 0 and a standard deviation of 1. It's the foundation for z-score calculations because once a score from any normal distribution is transformed into a z-score, it is expressed in terms of the standard normal distribution, regardless of the original mean or standard deviation.This distribution is critical because it allows us to use standard normal distribution tables (z-tables) to find probabilities associated with specific z-scores, which are otherwise not easily calculated without such tools. Since the properties of the standard normal distribution are well-known and tabulated, it serves as a reference framework for statistical analysis.

Importance of the Standard Normal Distribution

  • Universality: Any normal distribution can be converted to a standard normal distribution.
  • Comparison: It allows for the comparison of scores from different distributions.
  • Predictability: Patterns and probabilities are consistent and easily looked up due to standardization.
Statisticians and researchers highly value the standard normal distribution, often depicted by the bell curve, because of its predictability and the ease it brings to statistical analysis.
Probability Using Z-Table
A z-table lists the cumulative probabilities of the standard normal distribution up to a given z-score. It's an essential tool for finding the probability associated with a z-score without complex calculations. The table shows the probability that a standard normally distributed variable is less than or equal to a particular value.To use a z-table, you simply find the z-score for the value of interest (as detailed in the exercise solution), and then look up the corresponding probability in the z-table. If the z-score is positive, it refers to the area to the left of that z-score. If the z-score is negative, you must consider the symmetry of the normal distribution, and often you'll use the complement (1 minus the value) to find the probability of being greater than the value on the left side.

Using the Z-Table in the Exercise

For example, if we calculate a z-score for the paint mixture and find that our z-score is 2, we can look up this value in the z-table to find that approximately 97.72% of the data falls below this z-score. Conversely, if we are interested in calculating the probability that the amount of red dye is greater than a certain value, we use the complement of the z-table result, effectively finding the area under the curve to the right of the z-score.

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Most popular questions from this chapter

The distribution of the number of items produced by an assembly line during an 8 -hour shift can be approximated by a normal distribution with mean value 150 and standard deviation 10 . a. What is the approximate probability that the number of items produced is at most \(120 ?\) b. What is the approximate probability that at least 125 items are produced? c. What is the approximate probability that between 135 and 160 (inclusive) items are produced?

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