gpa\n2.46\n3.98\n1.85\n2.86\n1.88\n1.84\n2.19\n3.11\n1.75\n2.16\n3.81\n2.62\n3.1\n3.43\n2.74\n(a) find the…

Explanation:

Step1: Arrange data in ascending order

$1.75, 1.84, 1.85, 1.88, 2.16, 2.19, 2.46, 2.62, 2.74, 2.86, 3.1, 3.11, 3.43, 3.81, 3.98$

Step2: Calculate the position of first - quartile ($Q_1$)

$n = 15$, position of $Q_1=\frac{n + 1}{4}=\frac{15+1}{4}=4$. So $Q_1$ is the 4th value in the ordered list. $Q_1 = 1.88$

Step3: Calculate the position of third - quartile ($Q_3$)

Position of $Q_3=\frac{3(n + 1)}{4}=\frac{3\times(15 + 1)}{4}=12$. So $Q_3$ is the 12th value in the ordered list. $Q_3=3.11$

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

$\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{1.75+1.84+1.85+1.88+2.16+2.19+2.46+2.62+2.74+2.86+3.1+3.11+3.43+3.81+3.98}{15}=\frac{38.28}{15}=2.552$

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 $=2.62$

Step6: Calculate the range

Range $= \text{Max}-\text{Min}=3.98 - 1.75=2.23$

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

First, calculate the sum of squared deviations $\sum_{i=1}^{n}(x_i-\bar{x})^2$. $(1.75 - 2.552)^2+(1.84 - 2.552)^2+\cdots+(3.98 - 2.552)^2$ $=(- 0.802)^2+(-0.712)^2+\cdots+(1.428)^2$ $=0.643204 + 0.506944+\cdots+2.039184$ $=10.89972$. $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}=\sqrt{\frac{10.89972}{14}}\approx0.88$

Step8: Calculate the inter - quartile range (IQR)

$IQR=Q_3 - Q_1=3.11-1.88 = 1.23$

Step9: Calculate the lower and upper limits for outliers

Lower limit $LL=Q_1-1.5\times IQR=1.88-1.5\times1.23=1.88 - 1.845 = 0.035$ Upper limit $UL=Q_3+1.5\times IQR=3.11+1.5\times1.23=3.11+1.845 = 4.955$

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

$s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}=\frac{10.89972}{14}\approx0.78$

Answer:

(a) $1.88$ (b) $3.11$ (c) $2.552$ (d) $2.62$ (e) $2.23$ (f) $0.88$ (g) $1.23$ (h) $LL:0.035$, $UL:4.955$ (i) $0.78$