the following data represents the gpa of a sample of 15 students enrolled in a and round to 4 decimal…

Explanation:

Step1: Sort the data

1.7, 1.86, 2.08, 2.18, 2.25, 2.28, 2.44, 2.9, 2.95, 3.02, 3.05, 3.65, 3.7, 3.74, 3.91

Step2: Calculate the first quartile ($Q_1$)

The position of $Q_1$ for $n = 15$ data - points is $i=\frac{n + 1}{4}=\frac{15+1}{4}=4$. So, $Q_1$ is the 4th - ordered value. $Q_1 = 2.18$.

Step3: Calculate the third quartile ($Q_3$)

The position of $Q_3$ is $i = 3\times\frac{n + 1}{4}=3\times\frac{15 + 1}{4}=12$. So, $Q_3$ is the 12th - ordered value. $Q_3=3.65$.

Step4: Calculate the mean ($\bar{x}$)

$\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{1.7+2.25+2.18+2.08+3.91+2.95+3.65+1.86+2.44+3.7+3.74+2.9+2.28+3.05+3.02}{15}=\frac{41.79}{15}=2.7860$.

Step5: Calculate the median

Since $n = 15$ (odd), the median is the $\left(\frac{n + 1}{2}\right)$-th value. $\frac{15+1}{2}=8$th value. Median $=2.90$.

Step6: Calculate the range

Range $=\text{Max}-\text{Min}=3.91 - 1.7=2.21$.

Step7: Calculate the sample standard - deviation ($s$)

First, calculate the sum of squared deviations: $\sum_{i = 1}^{n}(x_i-\bar{x})^2=(1.7 - 2.7860)^2+(2.25 - 2.7860)^2+\cdots+(3.02 - 2.7860)^2$. $\sum_{i = 1}^{n}(x_i-\bar{x})^2 = 11.09794$. $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}=\sqrt{\frac{11.09794}{14}}\approx0.8897$.

Step8: Calculate the inter - quartile range (IQR)

$IQR=Q_3 - Q_1=3.65 - 2.18 = 1.47$.

Step9: Calculate the lower and upper fences

Lower fence $=Q_1-1.5\times IQR=2.18-1.5\times1.47=2.18 - 2.205=-0.025$. Upper fence $=Q_3 + 1.5\times IQR=3.65+1.5\times1.47=3.65 + 2.205=5.855$.

Step10: Calculate the sample variance ($s^2$)

$s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}=\frac{11.09794}{14}\approx0.7927$.

Answer:

(a) 2.1800 (b) 3.6500 (c) 2.7860 (d) 2.9000 (e) 2.2100 (f) 0.8897 (g) 1.4700 (h) Lower: - 0.0250, Upper: 5.8550 (i) 0.7927