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 X2in words. What is its probability? What is the probability thatX<2?

Short Answer

Expert verified

(a)P(X1)=0.9

(b)P(X2)=0.6

P(X<2)=0.3

Step by step solution

01

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” .

02

Explanation(part a)

The probability distribution is:

Value of X0
1
2
3
4
Probability0.1
0.2
0.3
0.3
0.1

Consider, Xbe the random variable that shows the number of nonword errors. The at least non-word error could be written as X1.

The probability is computed as:

P(X1)=1P(X<1)

=1P(X=0)

=10.1

=0.9

The probability is0.9.

03

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 eventsX2andX<2

04

Explanation(part b)

X2means that the non-word errors are 2or less.

The probabilities could be calculated as:

P(X2)=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.

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