Question

gpa
3
2.18
3.66
3.86
3.31
1.57
3.52
1.97
3.79
2.59
3.15
1.52
2.59
3.56
2.23
(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 interquartile range.
(h) what are the lower and upper limits to find outliers?
ll: ul:
(i) find s².

Explanation:

Step1: Sort the data

$1.52, 1.57, 1.97, 2.18, 2.23, 2.59, 2.59, 3, 3.15, 3.31, 3.52, 3.56, 3.66, 3.79, 3.86$

Step2: Calculate the first quartile ($Q_1$)

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

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 value in the sorted data, $Q_3=3.56$.

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

$\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{1.52+1.57+1.97+2.18+2.23+2.59+2.59+3+3.15+3.31+3.52+3.56+3.66+3.79+3.86}{15}=\frac{42.4}{15}\approx2.83$.

Step5: Calculate the median

Since $n = 15$ (odd), the median is the $(\frac{n + 1}{2})$ - th value. $\frac{15+1}{2}=8$ - th value, so the median is $3$.

Step6: Calculate the range

Range = Maximum - Minimum. Maximum = $3.86$, Minimum = $1.52$. Range=$3.86 - 1.52 = 2.34$.

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

First, calculate the sum of squared differences from the mean.
$\sum_{i = 1}^{n}(x_i-\bar{x})^2=(1.52 - 2.83)^2+(1.57 - 2.83)^2+\cdots+(3.86 - 2.83)^2$.
$\sum_{i = 1}^{n}(x_i-\bar{x})^2=(-1.31)^2+(-1.26)^2+\cdots+(1.03)^2$.
$\sum_{i = 1}^{n}(x_i-\bar{x})^2 = 1.7161+1.5876+\cdots+1.0609$.
$\sum_{i = 1}^{n}(x_i-\bar{x})^2=13.978$.
$s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}=\sqrt{\frac{13.978}{14}}\approx0.999\approx1.00$.

Step8: Calculate the inter - quartile range (IQR)

IQR=$Q_3 - Q_1=3.56 - 2.18 = 1.38$.

Step9: Calculate the lower and upper limits for outliers

Lower Limit (LL)=$Q_1-1.5\times IQR=2.18-1.5\times1.38=2.18 - 2.07 = 0.11$.
Upper Limit (UL)=$Q_3+1.5\times IQR=3.56+1.5\times1.38=3.56 + 2.07 = 5.63$.

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

$s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}=\frac{13.978}{14}\approx0.99986\approx1.00$.

Answer:

(a) $2.18$
(b) $3.56$
(c) $2.83$
(d) $3$
(e) $2.34$
(f) $1.00$
(g) $1.38$
(h) LL: $0.11$, UL: $5.63$
(i) $1.00$