Question

Spell-checking software catches “nonword errors,” which result in a string of letters that is not a word, as when “the” is typed as “teh.” When undergraduates are asked to write a 250-word essay (without spell-checking), the number X of nonword errors has the following distribution:

(a) Write the event “at least one nonword error” in terms of $X$. What is the probability of this event?

(b) Describe the event $X\le 2$in words. What is its probability? What is the probability that$X<2$?

Short answer

(a)$P(X\ge 1)=0.9$

(b)$P(X\le 2)=0.6$

$P(X<2)=0.3$

Step-by-step solution

Given information(part a)

Given in the question that, Spell-checking software catches “nonword errors,” which result in a string of letters that is not a word, as when “the” is typed as “teh.” When undergraduates are asked to write a 250-word essay (without spell-checking), the number X of nonword errors has the following distribution:

We need to find the probability of event “at least one nonword error” .

Explanation(part a)

The probability distribution is:

Value of $X$	$0$	$1$	$2$	$3$	$4$
Probability	$0.1$	$0.2$	$0.3$	$0.3$	$0.1$

Consider, $X$be the random variable that shows the number of nonword errors. The at least non-word error could be written as $X\ge 1$.

The probability is computed as:

$P(X\ge 1)=1-P(X<1)$

$=1-P(X=0)$

$=1-0.1$

$=0.9$

The probability is$0.9.$

Given information(part b)

Spell-checking software catches “nonword errors,” which result in a string of letters that is not a word, as when “the” is typed as “teh.” When undergraduates are asked to write a 250-word essay (without spell-checking), the number X of nonword errors has the following distribution:

We need to find the probability of events$X\le 2$and$X<2$

Explanation(part b)

$X\le 2$means that the non-word errors are $2$or less.

The probabilities could be calculated as:

$P(X\le 2)=P(X=0)+P(X=1)+P(X=2)$

$=0.1+0.2+0.3$

$=0.6$

Now,

$P(X<2)=P(X=0)+P(X=1)$

$=0.1+0.2$

$=0.3$

Thus, the probabilities are $0.6$ and $0.3$ respectively.