Question

Typographical and spelling errors can be either “nonword errors” or “word errors.” A nonword error is not a real word, as when “the” is typed as “teh.” A word error is a real word, but not the right word, as when “lose” is typed as “loose.” When students are asked to write a 250-word essay (without spell-checking), the number of nonword errors X in a randomly selected essay has mean 2.1 and standard deviation 1.136. The number of word errors Y in the essay has mean 1.0 and standard deviation 1.0. Calculate and interpret the mean of the sum S = X + Y.

Short answer

The Average mean value is$\mathrm{\mu }=3.1$

Step-by-step solution

Given information

The number of nonword errors X in a randomly selected essay has mean 2.1 and standard deviation 1.136

The number of word errors X in a randomly selected essay has mean 1.0 and standard deviation 1.0

$$\begin{gathered} {\mathrm{\mu }}_{\mathrm{X}}=2.1 \\ {\mathrm{\sigma }}_{\mathrm{X}}=1.136 \\ {\mathrm{\mu }}_{\mathrm{Y}}=1.0 \\ {\mathrm{\sigma }}_{\mathrm{Y}}=1.0 \end{gathered}$$

X = quantity of nonword errors in an essay chosen at random

Y= quantity of word errors in an essay chosen at random

Calculations

$$\begin{gathered} \mathrm{\mu }={\mathrm{\mu }}_{\mathrm{X}}+{\mathrm{\mu }}_{\mathrm{Y}} \\ =2.1+1.0 \\ =3.1 \end{gathered}$$

There are two different random variables with two different means,${\mathrm{\mu }}_{\mathrm{X}}=2.1$and ${\mathrm{\mu }}_{\mathrm{Y}}=1.0$

as a result, the average of the two random numbers ,the total of their means will be the variable.

On average, there will be a total number of 3.1 errors