Question

.Essay errors 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 has the following probability distribution:

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

$\mu X=2.1\sigma X=1.136$

Value of Y	0	1	2	3
Probability	0.4	0.3	0.2	0.1

$\mu Y=1.0\sigma Y=1.0$

a) Find the mean and standard deviation of the difference $Y-X$in the number of errors made by a randomly selected student. Interpret each value in context.

(b) Challenge: Find the probability that a randomly selected student makes more word errors than nonword errors .

Short answer

(a) $\sigma Y-X=1.513$

(b) The probability that a randomly selected student makes more word errors than nonword errors is$0.7673$

Step-by-step solution

Part (a) Step 1: Given Information

Given in the question that, the mean of the non word errors ${\mu }_X=2.1$, and standard deviation of the non word errors ${\sigma }_X=1.136$

Similarly, given mean of the word errors ${\mu }_Y=1.0$and standard deviation of the word errors ${\sigma }_Y=1.0$

We have to find the mean and standard deviation of the difference $Y-X$in the number of errors made by a randomly selected student.

Part (a) Step 2: Explanation 

We have to find the mean and standard deviation of$Y-X$

Therefore, the mean of $Y-X$,

localid="1649872702152" $$\begin{gathered} {\mu }_{Y-X}={\mu }_Y-{\mu }_X \\ =2.1-1.0 \\ =1.1 \end{gathered}$$

We can expect alocalid="1649872706557" $1.1$difference in the number of non-word and word errors on average.

So, the Standard deviation of $Y-X$,

localid="1649872711233" $$\begin{gathered} {\sigma }_{y-x}={\sqrt{{{\sigma }_y}^2+{\sigma }_x}}^2 \\ =\sqrt{{1.0}^2+{1.136}^2} \\ =\sqrt{2.290496} \\ =1.513. \end{gathered}$$

Part (b) Step 1:  Given Information 

We have to find the probability that a randomly selected student makes more word errors than nonword errors$P(Y-X\ge 0)$

Part (b) Step 2: Calculate the probability

Let's standardize $Y-X$$=0$

localid="1649872719676" $$\begin{gathered} z=\frac{x-{\mu }_{y-x}}{{\sigma }_{y-x}} \\ =\frac{0-1.1}{1.513} \\ =\frac{-1.1}{1.513} \\ =-0.73 \end{gathered}$$

Using a table of standard normal probability as an example:

localid="1649872722744" $$\begin{gathered} P(Y-X\ge 0)=P(z\ge -0.73) \\ =P(z\le 0.73) \\ =0.7673 \end{gathered}$$