gpa\n7.46\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: Sort the data

1.75, 1.84, 1.85, 1.88, 2.16, 2.19, 2.62, 2.74, 2.86, 3.1, 3.11, 3.43, 3.81, 3.98, 7.46

Step2: Calculate the position of the first - quartile

$n = 15$. The position of the first - quartile $Q_1$ is $i_1=\frac{1}{4}(n + 1)=\frac{1}{4}(15 + 1)=4$. So $Q_1$ is the 4th value in the sorted data, $Q_1 = 1.88$.

Step3: Calculate the position of the third - quartile

The position of the third - quartile $Q_3$ is $i_3=\frac{3}{4}(n + 1)=\frac{3}{4}(15 + 1)=12$. So $Q_3$ is the 12th value in the sorted data, $Q_3 = 3.43$.

Step4: Calculate the mean

$\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{7.46+3.98+1.85+2.86+1.88+1.84+2.19+3.11+1.75+2.16+3.81+2.62+3.1+3.43+2.74}{15}=\frac{40.28}{15}\approx2.6853$.

Step5: Calculate the median

Since $n = 15$ (odd), the median is the $\left(\frac{n + 1}{2}\right)$-th value. $\frac{15+1}{2}=8$, so the median is 2.74.

Step6: Calculate the range

Range = Max - Min. Max = 7.46, Min = 1.75. Range = 7.46 - 1.75 = 5.71.

Step7: Calculate the sample standard deviation $s$

First, calculate the sum of squared deviations from the mean. $\sum_{i = 1}^{n}(x_i-\bar{x})^2=(7.46 - 2.6853)^2+(3.98 - 2.6853)^2+\cdots+(2.74 - 2.6853)^2$. $\sum_{i = 1}^{n}(x_i-\bar{x})^2\approx31.977$. $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}=\sqrt{\frac{31.977}{14}}\approx1.51$.

Step8: Calculate the inter - quartile range (IQR)

IQR=$Q_3 - Q_1=3.43 - 1.88 = 1.55$.

Step9: Calculate the lower and upper limits for outliers

Lower Limit (LL)=$Q_1-1.5\times IQR=1.88-1.5\times1.55=1.88 - 2.325=-0.445$. Upper Limit (UL)=$Q_3 + 1.5\times IQR=3.43+1.5\times1.55=3.43+2.325 = 5.755$.

Step10: Calculate the sample variance $s^2$

$s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}=\frac{31.977}{14}\approx2.284$.

Answer:

(a) 1.88 (b) 3.43 (c) 2.6853 (d) 2.74 (e) 5.71 (f) 1.51 (g) 1.55 (h) LL: - 0.445, UL: 5.755 (i) 2.284