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

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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.

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