Question

gpa
1.7
2.75
2.18
2.08
3.93
2.95
3.65
1.86
2.44
3.7
3.74
2.9
2.28
3.05
3.02
(a) find the first quartile.
(b) find the third quartile.
(c) find the mean.
(d) find the median.
(e) find the range.
(f) find s.
(g) find the inter - quartile range.
(h) what are the lower and upper fences to find outliers?
lower:
upper:
(i) find s².

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 $s=\sqrt{s^{2}}$

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

Answer:

(c) Mean: $2.682$
(e) Range: $2.23$
(f) $s\approx0.7095$
(i) $s^{2}\approx0.5034$