/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 Consider the population of all o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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. \(P(4.6 < x < 5.2)\) e. \(P(x > 4.5)\) f. \(P(x > 4.0)\)

Short Answer

Expert verified
The respective probabilities are: a) 0.5; b) 0.9772; c) 0.9772; d) 0.8185; e) 0.9938; f) 1.0.

Step by step solution

01

Understand the Z-Score

In a normal distribution, the Z-score is the number of standard deviations a given data point is from the mean. It is calculated using the formula \(Z = \frac{x - μ}{σ}\), where x is the data point, μ is the mean and σ is the standard deviation.
02

Calculate a. \(P(x < 5.0)\)

First calculate the Z-score using the formula stated above: \(Z = \frac{x - μ}{σ} = \frac{5.0 - 5}{0.2} = 0\). As we are looking at a standard normal table, we see that \(P(Z < 0) = 0.5\), so \(P(x < 5.0) = 0.5\).
03

Calculate b. \(P(x < 5.4)\)

First determine the Z-score: \(Z = \frac{x - μ}{σ} = \frac{5.4 - 5}{0.2} = 2.0\). From standard normal table, we find that \(P(Z < 2.0) = 0.9772\), so \(P(x < 5.4) = 0.9772\).
04

Calculate c. \(P(x \leq 5.4)\)

This is the same as part b since in a continuous distribution, the probability at one specific point is indistinguishable. So \(P(x \leq 5.4) = 0.9772\).
05

Calculate d. \(P(4.6 < x < 5.2)\)

First calculate Z-scores for x = 4.6 and x = 5.2 to get \(Z_1 = -2.0\) and \(Z_2 = 1.0\). We then find the probabilities \(P(Z < -2.0) = 0.0228\) and \(P(Z < 1.0) = 0.8413\). The probability of x being between 4.6 and 5.2 ml is \(P(4.6 < x < 5.2) = P(Z < 1) - P(Z < -2) = 0.8413 - 0.0228 = 0.8185\).
06

Calculate e. \(P(x > 4.5)\)

First calculate the Z-score for 4.5, which gives us \(Z = -2.5\). Then find the probability corresponding to this Z-score: \(P(Z < -2.5) = 0.0062\). This gives the probability of being below 4.5, and as we want the probability of being greater than 4.5 we subtract it from 1: \(P(x > 4.5) = 1 - P(Z < -2.5) = 1 - 0.0062 = 0.9938\).
07

Calculate f. \(P(x > 4.0)\)

Calculate the Z-score for 4.0 which gives us \(Z = -5.0\). The standard normal table gives \(P(Z < -5) = 0\), hence \(P(x > 4.0) = 1 - P(Z < -5) = 1 - 0 = 1\). This indicates that it's almost certain to have more than 4 ml of red dye in the paint.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Z-score Calculation
The Z-score is a standard measure that quantifies the position of a data point relative to the mean of a group of data points in terms of standard deviations. It is a crucial concept for understanding the normal distribution and is calculated with the formula:
\[ Z = \frac{{x - \mu}}{{\sigma}} \]
Where:
  • \(x\) represents the value of the data point,
  • \(\mu\) is the mean of the distribution,
  • \(\sigma\) is the standard deviation.
In our example, to find out how many standard deviations a particular value of red dye (\(x\)) is from the mean (\(\mu\)), we use the above formula. For example, to find the Z-score for 5.0 ml of red dye, we calculate:
\[ Z = \frac{{5.0 - 5}}{{0.2}} = 0 \]
A Z-score of 0 indicates that the value is exactly at the mean. Z-scores also enable us to use standard normal distribution tables to find out probabilities associated with different values of \(x\) as seen in the steps of the provided solution.
Standard Deviation
Standard deviation (\(\sigma\)) is a statistical measure that describes the spread or dispersion of a set of values within a distribution. A larger standard deviation indicates that the values are more spread out from the mean. In the context of our exercise, a standard deviation of 0.2 ml implies that the amount of red dye in most one-gallon cans is fairly close to the mean. Intuitively, a small standard deviation in this scenario suggests that the paint company maintains consistent quality in their dye mixing process.
Calculating standard deviation is fundamental before you can work out Z-scores and probabilities for the normal distribution. As standard deviation represents consistency, understanding it can help you discern how reliable a set of data can be. When the standard deviation is known, as in our exercise, we can use it to calculate Z-scores, which in turn help us with determining probabilities.
Normal Distribution Model
The normal distribution model is a symmetric, bell-shaped curve that represents the spread of a dataset where most of the occurrences take place near the mean and fewer as you move away. This model is used to describe real-life values that cluster around a single mean value, like the amount of red dye in paint cans produced by a company.
The properties of a normal distribution include:
  • The mean, median, and mode of the distribution are all equal.
  • The curve is symmetric about the mean.
  • The area under the curve is 1, implying that the total probability is 100%.
  • Approximately 68% of the data falls within one standard deviation of the mean; 95% within two standard deviations; 99.7% within three.
In our exercise, we used the normal distribution model to calculate the probability of finding specific amounts of red dye in the paint cans, which enabled us to understand the quality control of the paint manufacturing process. Understanding the normal distribution model is essential to correctly interpret many natural and industrial processes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The probability distribution of \(x,\) the number of defective tires on a randomly selected automobile checked at a certain inspection station, is given in the following table: $$ \begin{array}{lccccc} x & 0 & 1 & 2 & 3 & 4 \\ p(x) & 0.54 & 0.16 & 0.06 & 0.04 & 0.20 \end{array} $$ a. Calculate the mean value of \(x\). b. Interpret the mean value of \(x\) in the context of a long sequence of observations of number of defective tires. c. What is the probability that \(x\) exceeds its mean value? d. Calculate the standard deviation of \(x\).

The probability distribution of \(x,\) the number of defective tires on a randomly selected automobile checked at a certain inspection station, is given in the following table: $$ \begin{array}{lccccc} x & 0 & 1 & 2 & 3 & 4 \\ p(x) & 0.54 & 0.16 & 0.06 & 0.04 & 0.20 \end{array} $$ The mean value of \(x\) is \(\mu_{x}=1.2\). Calculate the values of \(\sigma_{x}^{2}\) and \(\sigma_{x}\)

Suppose that the \(\mathrm{pH}\) of soil samples taken from a certain geographic region is normally distributed with a mean of 6.00 and a standard deviation of \(0.10 .\) Suppose the \(\mathrm{pH}\) of a randomly selected soil sample from this region will be determined. 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 \(\mathrm{pH}\) values?

Example 6.14 gave the probability distributions shown below for \(x=\) number of flaws in a randomly selected glass panel from Supplier 1 \(y=\) number of flaws in a randomly selected glass panel from Supplier 2 for two suppliers of glass used in the manufacture of flat screen TVs. If the manufacturer wanted to select a single supplier for glass panels, which of these two suppliers would you recommend? Justify your choice based on consideration of both center and variability. $$ \begin{array}{lcccclcccc} \boldsymbol{x} & 0 & 1 & 2 & 3 & \boldsymbol{y} & 0 & 1 & 2 & 3 \\ \boldsymbol{p}(\boldsymbol{x}) & 0.4 & 0.3 & 0.2 & 0.1 & \boldsymbol{p}(\boldsymbol{y}) & 0.2 & 0.6 & 0.2 & 0 \end{array} $$

Suppose that in a certain metropolitan area, \(90 \%\) of all households have cable TV. Let \(x\) denote the number among four randomly selected households that have cable TV. Then \(x\) is a binomial random variable with \(n=4\) and \(p=0.9\). a. Calculate \(p(2)=P(x=2)\), and interpret this probability. b. Calculate \(p(4)\), the probability that all four selected households have cable TV. c. Determine \(P(x \leq 3)\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.