/*! 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

Suppose \(y=\) the number of broken eggs in a randomly selected carton of one dozen eggs. The probability distribution of \(y\) is as follows: $$ \begin{array}{lccccl} y & 0 & 1 & 2 & 3 & 4 \\ p(y) & 0.65 & 0.20 & 0.10 & 0.04 & ? \end{array} $$ a. Only \(y\) values of \(0,1,2,3,\) and 4 have probabilities greater than 0 . What is \(p(4)\) ? b. How would you interpret \(p(1)=0.20 ?\) c. Calculate \(P(y \leq 2)\), the probability that the carton contains at most two broken eggs, and interpret this probability. d. Calculate \(P(y<2),\) the probability that the carton contains fewer than two broken eggs. Why is this smaller than the probability in Part (c)? e. What is the probability that the carton contains exactly 10 unbroken eggs? f. What is the probability that at least 10 eggs are unbroken?

Suppose that fuel efficiency (miles per gallon, mpg) for a particular car model under specified conditions is normally distributed with a mean value of \(30.0 \mathrm{mpg}\) and a standard deviation of \(1.2 \mathrm{mpg}\). a. What is the probability that the fuel efficiency for a randomly selected car of this model is between 29 and \(31 \mathrm{mpg} ?\) b. Would it surprise you to find that the efficiency of a randomly selected car of this model is less than \(25 \mathrm{mpg} ?\) c. If three cars of this model are randomly selected, what is the probability that each of the three have efficiencies exceeding \(32 \mathrm{mpg}\) ? d. Find a number \(x^{*}\) such that \(95 \%\) of all cars of this model have efficiencies exceeding \(x^{*}\left(\right.\) i.e., \(\left.P\left(x>x^{*}\right)=0.95\right)\).

Suppose that the distribution of typing speed in words per minute (wpm) for experienced typists using a new type of split keyboard can be approximated by a normal curve with mean 60 wpm and standard deviation 15 wpm ("The Effects of Split Keyboard Geometry on Upper Body Postures, Ergonomics [2009]: 104-111). a. What is the probability that a randomly selected typist's speed is at most 60 wpm? Less than 60 wpm? b. What is the probability that a randomly selected typist's speed is between 45 and 90 wpm? c. Would you be surprised to find a typist in this population whose speed exceeded 105 wpm? d. Suppose that two typists are independently selected. What is the probability that both their speeds exceed 75 wpm? e. Suppose that special training is to be made available to the slowest \(20 \%\) of the typists. What typing speeds would qualify individuals for this training? (Hint: See Example 6.23\()\)

You are to take a multiple-choice exam consisting of 100 questions with five possible responses to each question. Suppose that you have not studied and so must guess (randomly select one of the five answers) on each question. Let \(x\) represent the number of correct responses on the test. a. What kind of probability distribution does \(x\) have? b. What is your expected score on the exam? (Hint: Your expected score is the mean value of the \(x\) distribution.) c. Calculate the variance and standard deviation of \(x\). d. Based on your answers to Parts (b) and (c), is it likely that you would score over 50 on this exam? Explain the reasoning behind your answer.

Suppose \(y=\) the number of broken eggs in a randomly selected carton of one dozen eggs. The probability distribution of \(y\) is as follows: $$ \begin{array}{lccccc} y & 0 & 1 & 2 & 3 & 4 \\ p(y) & 0.65 & 0.20 & 0.10 & 0.04 & 0.01 \end{array} $$ a. Calculate and interpret \(\mu_{y}\). (Hint: See Example 6.13) b. In the long run, for what percentage of cartons is the number of broken eggs less than \(\mu_{y} ?\) Does this surprise you? c. Explain why \(\mu_{y}\) is not equal to \(\frac{0+1+2+3+4}{5}=2.0\).

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.