Chapter 6.2, Problem 62E

To determine

Standard deviation of the total score deduction T for a randomly selected essay.

Answer to Problem 62E

Standard deviation of the total score deduction,

  ${\sigma }_T\approx 3.9515$

Explanation of Solution

Given information:

X and Y are independent events.

In a randomly selected essay,

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,

  ${\sigma }_Y=1.0$

T : Total score deductions

For each nonword errors, there is a deduction of 3 points.

Thus,

The product of number of nonword errors and 3 gives the number of points deducted for nonword errors.

We know that

The standard deviation is a measure of spread.

When every data value is multiplied by 3, the measure of spread should also be multiplied by 3.

Thus,

Standard deviation for nonword errors,

  ${\sigma }_{\text{Nonword}}=3{\sigma }_X=3\left(1.136\right)=3.408$

Now,

For eachword errors, there is a deduction of 2 points.

Thus,

The product of number of word errorsand 2 gives the number of points deducted for word errors.

We know that

The standard deviation is a measure of spread.

When every data value is multiplied by 2, the measure of spread should also be multiplied by 2.

Thus,

Standard deviation for nonword errors,

  ${\sigma }_{\text{Word}}=2{\sigma }_Y=2\left(1.0\right)=2.0$

When random variables are independent, the variance of the sum of the two random variables equals the sum of their variances.

And

T represents the total score deductions.

  ${\sigma }_T^2={\sigma }_{\text{Nonword+Word}}^2={\sigma }_{\text{Nonword}}^2+{\sigma }_{\text{Word}}^2={\left(3.408\right)}^2+{\left(2.0\right)}^2=15.614464$

We know that

Standard deviation is the square root of the variance.

Thus,

  ${\sigma }_T={\sigma }_{\text{Main}-\text{Downtown}}=\sqrt{15.614464}\approx 3.9515$

Thus,

On an average, the total score deductions vary by 3.9515 from the mean deduction of 8.3.