Question

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 =

Answer 1

Answer:

a) X is binomial with n = 10 and p = 0.3

Y is binomial with n = 10 and p = 0.7

b) The mean number of errors caught is 7.

The mean number of errors missed is 3.

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

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

Step-by-step explanation:

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:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

10 word errors.

This means that $n = 10$

(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?

$E(Y) = np = 10*0.7 = 7$

The mean number of errors caught is 7.

What is the mean number of errors missed?

$E(X) = np = 10*0.3 = 3$

The mean number of errors missed is 3.

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

$\sqrt{V(Y)} = \sqrt{np(1-p)} = \sqrt{10*0.7*0.3} = 1.4491$

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

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

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{10*0.3*0.7} = 1.4491$

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