91Ó°ÊÓ

$X$: Number of nonword errors

$Y$: Number of word errors

Such that

For $X$:

Mean, ${\mathrm{μ}}_X=2.1$

Standard deviation, ${\mathrm{σ}}_X=1.136$

For $Y$:

Mean,${\mathrm{μ}}_Y=1.0$

Standard deviation,role="math" localid="1654186687339" ${\mathrm{σ}}_Y=1.0$

$T$: Total score deductions

There is a three-point deduction for each nonword error.

As a result, the number of points subtracted for nonword errors is the product of the number of nonword errors and $3$.

The standard deviation is a measure of spread, as we all know.

When every data value is multiplied by three, the spread measure should be multiplied by three as well.

Thus, ${\mathrm{σ}}_{Nonword}=3{\mathrm{σ}}_X=3(1.136)=3.408$

is the standard deviation for nonword mistakes.

There is now a two-point reduction for each word error.

As a result, the number of points lost for word errors is equal to the product of the number of word errors and $2$.

When every data value is multiplied by two, the spread measure should be multiplied by two as well.

As a result, the nonword error standard deviation is :

${\mathrm{σ}}_{Word}=2{\mathrm{σ}}_Y=2(1.0)=2.0.$

The variance of the sum of the two random variables equals the sum of their variances when the random variables are independent.

$T$, on the other hand, represents the entire score deductions.

role="math" localid="1654187015040" $$\begin{gathered} {{\mathrm{σ}}^2}_T={{\mathrm{σ}}^2}_{Nonword+Word} \\ ={{\mathrm{σ}}^2}_{Nonword}+{{\mathrm{σ}}^2}_{Word} \\ ={(3.408)}^2+{(2.0)}^2 \\ =15.614464 \end{gathered}$$

We know that the square root of the variance is the standard deviation.

$$\begin{gathered} {\mathrm{σ}}_T={\mathrm{σ}}_{Main-Downtown} \\ =\sqrt{15.614464} \\ ≈\mathbf{3}\mathbf{.}\mathbf{9515} \end{gathered}$$