91影视

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

Similarly, given mean of the word errors ${渭}_Y=1.0$and standard deviation of the word errors ${蟽}_Y=1.0$

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.

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

Therefore, the mean of $Y-X$,

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

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

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

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

localid="1649872719676" $$\begin{gathered} z=\frac{x-{渭}_{y-x}}{{蟽}_{y-x}} \\ =\frac{0鈭�1.1}{1.513} \\ =\frac{鈭�1.1}{1.513} \\ =鈭�0.73 \end{gathered}$$

Using a table of standard normal probability as an example:

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