Question

Quiz grades Joey’s first 14 quiz grades in a marking period were as follows:

(a) Calculate the mean. Show your work.

(b) Suppose Joey has an unexcused absence for the 15th quiz, and he receives a score of 0. Recalculate the mean. What property of the mean does this illustrate?

Short answer

Part (a) 85

Part (b) 79.333 and non-resistant property of mean.

Step-by-step solution

Part (a) Step 1. Given information.

The given data is shown below:

Part (a) Step 2. Find the mean.

$$\begin{gathered} n=14 \\ \overline{x}=\frac{\sum _{i=1}^n{x}_i}{n} \\ =\frac{\left(86+84+91+75+78+80+74+87+76+96+82+90+98+93\right)}{14} \\ =\frac{1190}{14} \\ =85 \end{gathered}$$

Part (b) Step 1. Find the mean including Joey's unexcused absence for the 15th quiz.

$$\begin{gathered} n=15 \\ \overline{x}=\frac{\sum _{i=1}^n{x}_i}{n} \\ =\frac{\left(86+84+91+75+78+80+74+87+76+96+82+90+98+93+0\right)}{15} \\ =\frac{1190}{15} \\ \approx 79.3333 \end{gathered}$$

This shows the property of mean named non-resistant property because the mean is influenced by the unexcused absence of Joy.