Question

janna is trying to analyze the grades shes earned so far this year. she already \translated\ the grades from their letters. 3, 3, 3.3, 2, 3.3, 3, 1, 3, 1, 1, 4, 4, 3.3, 3, 2 make a box - plot of these data.

Explanation:

Step1: Organize data in ascending - order

1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 3.3, 3.3, 3.3, 3.3, 4, 4

Step2: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $n = 18$ and $\sum_{i=1}^{18}x_{i}=1\times4 + 2\times2+3\times6 + 3.3\times4+4\times2=4 + 4+18+13.2 + 8=47.2$. So, $\bar{x}=\frac{47.2}{18}\approx2.62$

Step3: Calculate the standard - deviation

First, calculate the variance $s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}$.
$(1 - 2.62)^{2}\times4+(2 - 2.62)^{2}\times2+(3 - 2.62)^{2}\times6+(3.3 - 2.62)^{2}\times4+(4 - 2.62)^{2}\times2$
$=(- 1.62)^{2}\times4+(-0.62)^{2}\times2+(0.38)^{2}\times6+(0.68)^{2}\times4+(1.38)^{2}\times2$
$=2.6244\times4 + 0.3844\times2+0.1444\times6+0.4624\times4+1.9044\times2$
$=10.4976+0.7688 + 0.8664+1.8496+3.8088=17.7912$
$s^{2}=\frac{17.7912}{17}\approx1.0465$, and $s=\sqrt{1.0465}\approx1.023$

Step4: Find the minimum

The minimum value, Min = 1

Step5: Find the first quartile (Q1)

Since $n = 18$, the position of Q1 is $\frac{n + 1}{4}=4.75$. So, Q1 is the value between the 4th and 5th ordered data points. Q1 = 1.5

Step6: Find the median

Since $n = 18$, the median is the average of the 9th and 10th ordered data points. Median=$\frac{3 + 3}{2}=3$

Step7: Find the third quartile (Q3)

The position of Q3 is $\frac{3(n + 1)}{4}=14.25$. So, Q3 is the value between the 14th and 15th ordered data points. Q3 = 3.3

Step8: Find the maximum

The maximum value, Max = 4

Answer:

	Value
St. Dev	1.023
Min	1
Q1	1.5
Median	3
Q3	3.3
Max	4