Spell Checking Software Catches Nonword Errors
Spell Checking Software Catches Nonword Errors Which Are Strings Of
Spell-checking software catches "nonword errors," which are strings of letters that are not words, such as when "the" is typed as "eth." When undergraduates are asked to write a 250-word essay without spell-checking, the number X of nonword errors follows a certain probability distribution. The tasks involve sketching this probability distribution, expressing and calculating probabilities of specific events related to X, and interpreting these events in words.
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Nonword errors in student writing, particularly when spell-checking tools are disabled, provide a compelling case study for understanding probability distributions and event probabilities. The distribution of the total number of nonword errors (X) in a 250-word essay can be modeled using discrete probability models—most likely a binomial or Poisson distribution, depending on the context. The problem requires sketching the probability distribution, which involves plotting the probability mass function for X over its possible range, typically from zero errors upward.
Specifically, the probability distribution for X can be visualized as a series of discrete points, where the height of each point corresponds to the probability of exactly that number of errors occurring. For instance, the probability that X equals 0 (no errors) might be the highest, with probabilities decreasing as X increases, depending on the errors' likelihood per word. The shape of the distribution reveals the tendency of students to make a certain number of errors and can be characterized as either skewed or approximately symmetric, based on the underlying error rate.
Understanding the events involving X is crucial for interpreting the error likelihood. The event "at least one nonword error" can be expressed as X ≥ 1, which encompasses all instances where one or more errors are present. Conversely, the event "X ≥ 2" refers to two or more errors occurring in the essay. The probability of these events can be calculated from the distribution, with P(X ≥ 1) = 1 - P(X = 0), which gives the probability of having at least one error.
Further, describing the event X ≤ 2 in words involves stating that the number of nonword errors is no more than two, meaning that the errors are rare or minimal. The probability associated with this event includes the probabilities of X being 0, 1, or 2, summed together. The complement, or the probability that X
Moving beyond this, the distribution model helps in understanding error patterns and setting expectations for error correction strategies in educational contexts. These insights inform assessments of writing quality and the efficacy of pedagogical tools aimed at reducing spelling mistakes.
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In the second scenario, Sheila's glucose levels are studied to determine the likelihood of gestational diabetes. The measured glucose levels, modeled as a normal distribution with mean μ = 128 mg/dl and standard deviation σ = 10 mg/dl, are analyzed to find the value L such that the probability of any test result exceeding L is only 0.05. This involves using the properties of the normal distribution, specifically the z-score, to find the critical value corresponding to the upper 5% tail.
Calculating this critical level involves determining the z-score associated with a probability of 0.95 (since we want the upper tail to be 0.05). From standard normal distribution tables or computation, the z-score for 0.95 is approximately 1.645. Applying the z-score formula:
L = μ + z σ = 128 + 1.645 10 ≈ 128 + 16.45 ≈ 144.5 mg/dl.
Thus, there is a 95% probability that the mean glucose level of two tests will fall below approximately 144.5 mg/dl, providing a threshold for concern in diagnosis.
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Benford's Law presents a fascinating statistical pattern that predicts the distribution of first digits in naturally occurring datasets. The probability that a number's first digit is d (1 through 9) according to Benford's Law is given by:
P(d) = log10(1 + 1/d)
Applying this, the probability of the first digit being 1, 2, or 3 can be summed to find the likelihood that a randomly selected invoice has a first digit in this range. Specifically:
P(1) = log10(2) ≈ 0.3010
P(2) = log10(1 + 1/2) ≈ 0.1761
P(3) = log10(1 + 1/3) ≈ 0.1249
Sum: ≈ 0.6020
Using the binomial distribution to model the number of invoices in a sample of 1234 with first digits 1, 2, or 3, the probability that at most 678 such invoices occur can be approximated using normal approximation to the binomial distribution. The mean number of invoices with these first digits is:
μ = 1234 * 0.6020 ≈ 743.0
and the standard deviation:
σ = sqrt(1234 0.6020 (1 - 0.6020)) ≈ sqrt(1234 0.6020 0.3980) ≈ 15.4
The probability that the number of such invoices is less than or equal to 678 can then be evaluated using the normal distribution approximation, calculating the z-score:
Z = (678 + 0.5 - μ) / σ ≈ (678.5 - 743.0) / 15.4 ≈ -4.163
which corresponds to a cumulative probability close to zero, indicating it's highly unlikely under the assumption of Benford's distribution. Precise calculation yields an approximate probability of this event.
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Next, we analyze the weight of eggs produced by hens, which follows a normal distribution with mean μ = 67.2 grams and standard deviation σ = 6.9 grams. To find the probability that the total weight of a carton of 12 eggs falls between 775 g and 825 g, we first determine the total mean and standard deviation of the summed weights:
Mean total weight: μ_total = 12 * 67.2 = 806.4 g
Standard deviation of total: σ_total = sqrt(12) 6.9 ≈ 2.828 6.9 ≈ 19.54 g
Applying the normal distribution, we calculate the z-scores for 775 g and 825 g:
Z lower = (775 - 806.4) / 19.54 ≈ -1.599
Z upper = (825 - 806.4) / 19.54 ≈ 0.94
The probability that total weight P lies between 775 g and 825 g is then:
P = P(Z
Using standard normal distribution tables or software, P(Z
P ≈ 0.8264 - 0.0554 ≈ 0.7710.
This suggests a high probability that a randomly selected carton of 12 eggs will weigh within this specified range.
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Finally, the scenario involving the agency's mailing list concerns the probability of selecting 30 individuals with a certain gender composition or order. The probability that exactly 19 out of 30 are men, assuming each selection is independent and the probability of choosing a male is 0.65, follows the binomial distribution:
P(X=19) = C(30, 19) (0.65)19 (0.35)11
Using a calculator or statistical software, this yields a probability rounded to four decimal places.
Additionally, the probability that the first woman is reached on the 5th call involves the sequence of the first four calls being men (each with probability 0.65), and the fifth being a woman (probability 0.35), expressed as:
P = (0.65)4 * 0.35
This type of probability uses the geometric distribution principle, modeling the number of trials until the first success (here, finding a woman). Calculating yields a specific probability rounded accordingly.
References
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