/*! 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 5 Proofreading. Typing errors in a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Proofreading. Typing errors in a text are either nonword errors (as when "the" is typed as "teh") or word errors that result in a real but incorrect word. Spellchecking software will catch nonword errors but not word errors. Human proofreaders catch \(70 \%\) of word errors. You ask a fellow student to proofread an essay in which you have deliberately made 10 word errors. (a) If the student matches the usual \(70 \%\) rate, what is the distribution of the number of errors caught? What is the distribution of the number of errors missed? (b) Missing 3 or more out of 10 errors seems a poor performance. What is the probability that a proofreader who catches \(70 \%\) of word errors misses exactly 3 out of 10 ? If you use software, also find the probability of missing 3 or more out of 10 .

Short Answer

Expert verified
(a) Errors caught: Binomial(10, 0.7); missed: Binomial(10, 0.3). (b) Probability of missing exactly 3: ~0.267; 3 or more: ~0.617.

Step by step solution

01

Define Random Variables

Let's denote the random variable \(X\) as the number of word errors caught by the student. Since the student catches \(70\%\) of word errors, \(X\) is a binomial random variable with parameters \(n = 10\) and \(p = 0.7\).
02

Distribution of Errors Caught

The distribution of the number of errors caught by the student, \(X\), follows a binomial distribution: \(X \sim \text{Binomial}(n = 10, p = 0.7)\). This means the probability of catching \(k\) errors is given by \(P(X = k) = \binom{10}{k} (0.7)^k (0.3)^{10-k}\).
03

Distribution of Errors Missed

The number of errors missed, \(Y\), is the complement of \(X\). Thus, \(Y = 10 - X\), and \(Y\) also follows a binomial distribution with parameters \(n = 10\) and \(p = 0.3\); so, \(Y \sim \text{Binomial}(n = 10, p = 0.3)\).
04

Probability of Missing 3 Errors

To find the probability of missing exactly 3 out of 10 errors, calculate \(P(Y = 3)\). Using the binomial probability formula: \[ P(Y = 3) = \binom{10}{3} (0.3)^3 (0.7)^{7} \].
05

Calculate Probability of Missing 3 Errors

Compute \(P(Y = 3)\) as follows: \[ P(Y = 3) = \binom{10}{3} (0.3)^3 (0.7)^{7} = 120 \times 0.027 \times 0.0823543 \approx 0.2668 \].
06

Probability of Missing 3 or More Errors

The probability of missing 3 or more errors is \(P(Y \geq 3)\). This is calculated as: \[ P(Y \geq 3) = 1 - P(Y < 3) = 1 - [P(Y = 0) + P(Y = 1) + P(Y = 2)] \].
07

Calculate Probability of Missing Less Than 3 Errors

Compute \(P(Y = 0)\), \(P(Y = 1)\), and \(P(Y = 2)\) using the binomial formula and sum them up. - \(P(Y = 0) = \binom{10}{0} (0.3)^0 (0.7)^{10} = 0.0282\)- \(P(Y = 1) = \binom{10}{1} (0.3)^1 (0.7)^{9} = 0.1211\)- \(P(Y = 2) = \binom{10}{2} (0.3)^2 (0.7)^{8} = 0.2335\).Thus, \(P(Y < 3) = 0.0282 + 0.1211 + 0.2335 = 0.3828\).
08

Final Probability Calculation

Now, calculate \(P(Y \geq 3)\) as: \[ P(Y \geq 3) = 1 - 0.3828 = 0.6172 \].

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.

Probability Calculation
Probability is at the heart of this exercise. We are tasked with understanding how likely certain events are to happen, such as a student catching or missing typing errors in an essay. The scenario described uses a type of probability known as *binomial distribution*.
This distribution is helpful when dealing with situations that have a fixed number of trials, where each trial has only two possible outcomes, like success or failure.
  • In this context, a "success" means the student correctly catches an error.
  • The probability of success (catching an error) in each trial is 0.7, since 70% of errors are expected to be caught.
The number of errors caught (X) follows a binomial distribution, represented mathematically as: \(P(X = k) = \binom{10}{k} (0.7)^k (0.3)^{10-k}\).
Similarly, the number of errors missed (Y) also follows a binomial distribution but with the probability of failure: \(P(Y = j) = \binom{10}{j} (0.3)^j (0.7)^{10-j}\). Understanding these probabilities helps us calculate how often certain numbers of errors might be caught or missed.
Random Variables
In probability and statistical analysis, a random variable is a variable whose possible values depend on the outcome of a random process. In this problem, two random variables are defined: X and Y.
The random variable X represents the number of errors caught by the proofreader. It is governed by the binomial distribution since each error is a separate trial with a 70% chance of being caught. This means that X could take any integer value between 0 and 10, depending on how many errors are caught.
  • X is said to be binomially distributed with parameters n=10 (number of trials) and p=0.7 (probability of catching an error).
On the other hand, the random variable Y symbolizes the number of errors missed out of 10. Since Y is the 'complement' of X, when the student misses an error, it's because they didn't catch it. Consequently, the probability of missing an error is the opposite of catching one, or 0.3.
Random variables allow us to use mathematical formulas to predict outcomes and understand the events in probabilistic terms, which is essential in determining how many errors might be caught or missed.
Proofreading Errors
Understanding the nature of proofreading errors is crucial in forming realistic expectations about the proofreading process. Proofreading is the final step in the editing process where errors are caught and fixed, typically involving spelling, grammar, and typographical issues.
In this scenario, we are particularly focused on two types of errors:
  • *Nonword errors*, where the word typed doesn't exist in the dictionary, and software can catch them.
  • *Word errors*, where an incorrect real word is used, and these are harder for software to catch, requiring human intervention.
The exercise addresses the latter since human proofreaders are crucial to catching word errors at a 70% efficiency rate.
In our problem, you deliberately insert 10 errors into the essay. The probability model helps determine how many of those errors the proofreader is likely to miss because the model accounts for natural human error rates.
This approach to proofreading accounts for the variability innate in human performance and provides expectations on the proofreading efficacy, quantifying both detection and oversight possibilities.

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

Working Cell Numbers. When an opinion poll selects cell phone numbers at random to dial, the cell phone exchange is first selected and then random digits are added to form a complete telephone number (see Example 8.5). When using this procedure to generate random cell phone numbers, approximately \(55 \%\) of the cell numbers generated correspond to working numbers. You watch a pollster dial 15 cell numbers that have been selected in this manner. (a) What is the mean number of calls that reach working cell number? (b) What is the standard deviation \(\sigma\) of the count of calls that reach a working cell number? (c) Suppose that the probability of reaching a working cell number was \(p=0.70\). How does this new \(p\) affect the standard deviation? What would be the standard deviation if \(p=0.80\) ? What does your work show about the behavior of the standard deviation of a binomial distribution as the probability of a success gets closer to one?

Larry reads that one out of four eggs contains salmonella bacteria. So he never uses more than three eggs in cooking. If eggs do or don't contain salmonella independent of each other, the number of contaminated eggs when Larry uses three chosen at random has the distribution (a) binomial with \(n=4\) and \(p=1 / 4\). (b) binomial with \(n=3\) and \(p=1 / 4\). (c) binomial with \(n=3\) and \(p=1 / 3\).

In a group of 10 college students, 4 are business majors. You choose 3 of the 10 students at random and ask their major. The distribution of the number of business majors you choose is (a) binomial with \(n=10\) and \(p=0.4\). (b) binomial with \(n=3\) and \(p=0.4\). (c) not bìnomial.

Checking for Survey Errors. One way of checking the effect of undercoverage, nonresponse, and other sources of error in a sample survey is to compare the sample with known facts about the population. About \(22 \%\) of the Canadian population is first generation, that is, they were bom outside Canada. \({ }^{6}\) The number \(X\) of first-generation Canadians in random samples of 1500 persons should therefore vary with the binomial ( \(n=1500, p=0.22\) ) distribution. (a) What are the mean and standard deviation of \(X\) ? (b) Use the Nomal approximation to find the probability that the sample will contain between 340 and 390 first-generation Canadians. Be sure to check that you can safely use the approximation.

Antibiotic resistance. According to CDC estimates, more than 2 million people in the United States are sickened each year with antibiotic-resistant infections, with at least 23,000 dying as a result. Antibiotic resistance occurs when diseasecausing microbes become resistant to antibiotic drug therapy. Because this resistance is typically genetic and transferred to the next generations of microbes, it is a very serious public health problem. Of the three infections considered most serious by the CDC, gonorrhea has an estimated 800,000 cases occurring annually, with approximately \(30 \%\) of those cases resistant to any antibiotic \({ }^{7}\) A public health clinic in California sees eight patients with gonorrhea in a given week. (a) What is the distribution of \(X\), the number of these eight cases that are resistant to any antibiotic? (b) What are the mean and standard deviation of \(X\) ? (c) Find the probability that exactly one of the cases is resistant to any antibiotic. What is the probability that at least one case is resistant to any antibiotic? (Hint: It is easier to first find the probability that exactly zero of the eight cases were resistant.)

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.