Question

Essay scores : Refer to Exercise 50.

Assume that the number of nonword errors $X$ and word errors $Y$ in a randomly selected essay are independent random variables. An English professor deducts $3$ points from a student’s essay score for each nonword error and $2$ points for each word error. Find the standard deviation of the total score deductions $T$ for a randomly selected essay.

Short answer

Total score deductions vary by $\mathbf{3}\mathbf{.}\mathbf{9515}$ on average from the mean deduction of $8.3$.

Step-by-step solution

Given Information 

$X$: Number of nonword errors

$Y$: Number of word errors

Such that

For $X$:

Mean, ${\mu }_X=2.1$

Standard deviation, ${\sigma }_X=1.136$

For $Y$:

Mean,${\mu }_Y=1.0$

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

$T$: Total score deductions

Calculating standard deviation of the nonword errors

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, ${\sigma }_{Nonword}=3{\sigma }_X=3(1.136)=3.408$

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 $2$.

Find the standard deviation of the total score deductions T for a randomly selected essay. 

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 :

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

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" $$\begin{gathered} {{\sigma }^2}_T={{\sigma }^2}_{Nonword+Word} \\ ={{\sigma }^2}_{Nonword}+{{\sigma }^2}_{Word} \\ ={(3.408)}^2+{(2.0)}^2 \\ =15.614464 \end{gathered}$$

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

$$\begin{gathered} {\sigma }_T={\sigma }_{Main-Downtown} \\ =\sqrt{15.614464} \\ \approx \mathbf{3}\mathbf{.}\mathbf{9515} \end{gathered}$$