Question

Curving Test Scores A professor gives a test and the scores are normally distributed with a mean of 60 and a standard deviation of 12. She plans to curve the scores.

a. If she curves by adding 15 to each grade, what is the new mean and standard deviation?

b. Is it fair to curve by adding 15 to each grade? Why or why not?

c. If the grades are curved so that grades of B are given to scores above the bottom 70% and below the top 10%, find the numerical limits for a grade of B.

d. Which method of curving the grades is fairer: adding 15 to each original score or using a scheme like the one given in part (c)? Explain.

Short answer

a. The new mean and the new standard deviation after adding 15 to each grade are 75 and 12, respectively.

b. No, it is not fair to curve by adding 15 to each grade because the variation in the grades remains unchanged.

c. The numerical limits for grade B are 66.36 to 75.48.

d. The method used in part (c) is fairer because this method includes the change in the variation present in the original scores.

Step-by-step solution

Given information

Test scores are normally distributed with a mean $\mu =60$ and a standard deviation $\sigma =12$.

Describe the random variable

Let X be the test scores.

X follows a normal distribution with a mean equal to 60 and a standard deviation equal to 12.

Thus, $\mu$ is equal to 60 and $\sigma$ is equal to 12.

Find the new mean and standard deviation

a.

The mean is the sum of all the values divided by the number of observations.

If 15is added to every grade, then the sum of all the grades increases by 15 times n. Here, n is the number of grades available.

$$\begin{gathered} {\mu }_{new}=\frac{\text{Original}\text{Sum}+15n}{n} \\ =\frac{\text{Original}\text{Sum}}{n}+\frac{15n}{n} \\ =\mu +15 \\ =60+15 \\ =75 \end{gathered}$$

Thus, the new mean is equal to 75.

The standard deviation measures the amount of variation in the data set. If a constant value is added to every observation, then the variation in the data set remains unchanged, and thus the standard deviation remains the same.

In other words, the standard deviation is independent of change of origin.

Therefore,

$$\begin{gathered} {\sigma }_{\text{new}}=\sigma \\ =12 \end{gathered}$$

Thus, the new mean is equal to 75, and the new standard deviation is equal to 12.

Impact of curving the score by adding 15 to every grade

b.

By adding 15 to every grade, the average score increases, but the variation in the scores remains the same.

It would have been justified if both the average score and the variation in the scores had changed.

Thus, it is not fair to curve the scores by adding 15 to each grade because the variation remains unchanged.

Find the numerical limits for grade B

c.

The grades are curved so that grade B is given to scores above the bottom 70% and below the top 10%.

The score above the bottom 70% can be written as follows:

$P(Z<z)=0.70$

From the standard normal table, the area of 0.70 is observed corresponding to the row value of 0.5 and column value of 0.03. Thus, the z-score is equal to 0.53.

The scores below the top 10% can be written as follows:

$$\begin{gathered} P\left(Z>z\right)=0.10 \\ P\left(Z<z\right)=1-0.10 \\ =0.90 \end{gathered}$$

From the standard normal table, the area of 0.90 is observed corresponding to the row value of 1.2 and the column value of 0.09. Thus, the z-score is 1.29.

The conversion of the z-score to the sample value has the following expression:

$\text{x}=\mu +\text{z}\times \sigma$

Hence, for the z-score equal to 0.53, the sample value of the grade is equal to:

$$\begin{gathered} \text{x}=\mu +\text{z}\times \sigma \\ =60+0.53\times 12 \\ =66.36 \end{gathered}$$

Hence, for the z-score equal to 1.29, the sample value of the grade is equal to:

$$\begin{gathered} \text{x}=\mu +\text{z}\times \sigma \\ =60+1.29\times 12 \\ =75.48 \end{gathered}$$

Therefore, the numerical limits for grade B are66.36 and 75.48.

Discuss which method of curving the grades is fairer

d.

The method used in part (c) is fairer than the method, which involvesadding 15 to each original score because the method used in part (c)accounts for the variation in the scores and appears to be just.