gpa\n1.7\n2.75\n2.18\n2.08\n3.93\n2.95\n3.65\n1.86\n2.44\n3.7\n3.74\n2.9\n2.28\n3.05\n3.02\n(a) find the…

gpa\n1.7\n2.75\n2.18\n2.08\n3.93\n2.95\n3.65\n1.86\n2.44\n3.7\n3.74\n2.9\n2.28\n3.05\n3.02\n(a) find the first quartile.\n(b) find the third quartile.\n(c) find the mean.\n(d) find the median.\n(e) find the range.\n(f) find s.\n(g) find the inter - quartile range.\n(h) what are the lower and upper fences to find outliers?\nlower:\nupper:\n(i) find s².

Answer

Explanation:

Step1: Organize data in ascending order

$1.7, 1.86, 2.08, 2.18, 2.28, 2.44, 2.75, 2.9, 2.95, 3.02, 3.05, 3.65, 3.7, 3.74, 3.93$

Step2: Calculate the mean formula $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$

$n = 15$, $\sum_{i=1}^{15}x_{i}=1.7 + 1.86+2.08+2.18+2.28+2.44+2.75+2.9+2.95+3.02+3.05+3.65+3.7+3.74+3.93 = 40.23$ $\bar{x}=\frac{40.23}{15}=2.682$

Step3: Calculate the range formula $R=\max(x_{i})-\min(x_{i})$

$\max(x_{i}) = 3.93$, $\min(x_{i})=1.7$ $R=3.93 - 1.7=2.23$

Step4: Calculate the sample - variance formula $s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}$

First, calculate $(x_{i}-\bar{x})^{2}$ for each $i$: $(1.7 - 2.682)^{2}=(- 0.982)^{2}=0.964324$ $(1.86 - 2.682)^{2}=(-0.822)^{2}=0.675684$ $\cdots$ $(3.93 - 2.682)^{2}=(1.248)^{2}=1.557504$ $\sum_{i = 1}^{15}(x_{i}-\bar{x})^{2}=7.04796$ $s^{2}=\frac{7.04796}{14}\approx0.5034$

Step5: Calculate the sample - standard deviation formula $s=\sqrt{s^{2}}$

$s=\sqrt{0.5034}\approx0.7095$