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. Spell‑checking 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 X is the number of word errors missed, what is the distribution of X ? Select an answer choice. X is binomial with n = 10 and p = 0.7 X is binomial with n = 10 and p = 0.3 X is approximately Normal with μ = 3 and σ = 1.45 X is Normal with μ = 7 and σ = 1.45 If Y is the number of word errors caught, what is the distribution of Y ? Select an answer choice. Y is Normal with μ = 7 and σ = 1.45 Y is approximately Normal with μ = 3 and σ = 1.45 Y is binomial with n = 10 and p = 0.3 Y is binomial with n = 10 and p = 0.7 (b) What is the mean number of errors caught? (Enter your answer as a whole number.) mean of errors caught = What is the mean number of errors missed? (Enter your answer as a whole number.) mean of errors missed = (c) What is the standard deviation of the number of errors caught? (Enter your answer rounded to four decimal places.) standard deviation of the number of errors caught = What is the standard deviation of the number of errors missed? (Enter your answer rounded to four decimal places.) standard deviation of the number of errors missed =

For each typing error, there are only two possible outcomes. Either it is caught, or it is not. The probability of a typing error being caught is independent of other errors. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

10 word errors.

This means that [tex]n = 10[/tex]

(a) If X is the number of word errors missed, what is the distribution of X ?

Human proofreaders catch 70 % of word errors. This means that they miss 30% of errors.

So for X, p = 0.3.

The answer is:

X is binomial with n = 10 and p = 0.3.

If Y is the number of word errors caught, what is the distribution of Y ?

Human proofreaders catch 70 % of word errors.

So for Y, p = 0.7.

The answer is:

Y is binomial with n = 10 and p = 0.7

(b) What is the mean number of errors caught?

[tex]E(Y) = np = 10*0.7 = 7[/tex]

The mean number of errors caught is 7.

What is the mean number of errors missed?

[tex]E(X) = np = 10*0.3 = 3[/tex]

The mean number of errors missed is 3.

(c) What is the standard deviation of the number of errors caught?

[tex]\sqrt{V(Y)} = \sqrt{np(1-p)} = \sqrt{10*0.7*0.3} = 1.4491[/tex]

The standard deviation of the number of errors caught is 1.4491.

What is the standard deviation of the number of errors missed?

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{10*0.3*0.7} = 1.4491[/tex]

The standard deviation of the number of errors missed is 1.4491.