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Question
monday worksheet - test scores
class a test scores (out of 100)
student score
1 88
2 92
3 75
4 85
5 90
6 78
7 84
8 95
9 80
10 89
tasks: 1. find the mean, median, mode, and range. 2. calculate the mad. 3. estimate the standard deviation. 4. create a dot plot and a box plot of the data. 5. write 2 - 3 sentences interpreting the graphs: what do they tell you about the class scores?
Step1: Calculate the mean
$\text{Mean}=\frac{88 + 92+75+85+90+78+84+95+80+89}{10}=\frac{856}{10}=85.6$
Step2: Calculate the median
First, order the data: $75,78,80,84,85,88,89,90,92,95$. Since $n = 10$ (even), $\text{Median}=\frac{85 + 88}{2}=86.5$
Step3: Find the mode
No score repeats, so $\text{Mode}=\text{None}$
Step4: Calculate the range
$\text{Range}=\text{Max}-\text{Min}=95 - 75=20$
Step5: Calculate the MAD
- Find the absolute - deviations from the mean:
$|88 - 85.6| = 2.4$, $|92 - 85.6| = 6.4$, $|75 - 85.6| = 10.6$, $|85 - 85.6| = 0.6$, $|90 - 85.6| = 4.4$, $|78 - 85.6| = 7.6$, $|84 - 85.6| = 1.6$, $|95 - 85.6| = 9.4$, $|80 - 85.6| = 5.6$, $|89 - 85.6| = 3.4$
- Calculate the mean of the absolute - deviations:
$\text{MAD}=\frac{2.4+6.4 + 10.6+0.6+4.4+7.6+1.6+9.4+5.6+3.4}{10}=\frac{52}{10}=5.2$
Step6: Estimate the standard deviation
A rough estimate of the standard deviation $s\approx1.25\times\text{MAD}$. So $s\approx1.25\times5.2 = 6.5$
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- Mean: $85.6$, Median: $86.5$, Mode: None, Range: $20$
- MAD: $5.2$
- Estimated Standard Deviation: $6.5$
- (Dot - plot and box - plot creation is not shown in text format, but for a dot - plot, you would place a dot above each score value on a number line. For a box - plot, you would mark the minimum ($75$), first quartile ($Q_1 = 80$), median ($86.5$), third quartile ($Q_3 = 90$), and maximum ($95$))
- The dot plot shows the individual scores and their distribution. The box plot shows that the middle 50% of the scores are between $80$ and $90$, with a median of $86.5$. The spread of the scores as shown by the range and standard deviation indicates that while most scores are relatively close to the mean, there is some variability in the class test scores.