Question

3.65
1.86
2.44
3.7
3.74
2.9
2.28
3.05
3.02
(a) find the first quartile. 2.1800
(b) find the third quartile. 3.6500
(c) find the mean. 2.682
(d) find the median. 2.9000
(e) find the range. 2.23
(f) find s. 0.7095
(g) find the interquartile range. 1.4700
(h) what are the lower and upper fences to find outliers? lower: 0.0250 upper: 5.8550
(i) find s². 0.5034

Explanation:

Step1: Arrange data in ascending order

$1.86, 2.28, 2.44, 2.9, 3.02, 3.05, 3.65, 3.7, 3.74$

Step2: Calculate first quartile ($Q_1$)

The position of $Q_1$ is $0.25\times(n + 1)$ where $n=9$. So, $0.25\times(9 + 1)=2.5$. The value of $Q_1$ is the average of the 2nd and 3rd ordered - values. $Q_1=\frac{2.28 + 2.44}{2}=2.36$

Step3: Calculate third quartile ($Q_3$)

The position of $Q_3$ is $0.75\times(n + 1)=0.75\times(9 + 1)=7.5$. The value of $Q_3$ is the average of the 7th and 8th ordered - values. $Q_3=\frac{3.65+3.7}{2}=3.675$

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

$\bar{x}=\frac{1.86 + 2.28+2.44 + 2.9+3.02+3.05+3.65+3.7+3.74}{9}=\frac{24.69}{9}=2.7433\cdots\approx2.743$

Step5: Calculate the median

Since $n = 9$ (odd), the median is the 5th ordered - value, which is $3.02$

Step6: Calculate the range

Range = Maximum value - Minimum value. Maximum value is $3.74$ and minimum value is $1.86$. Range $=3.74−1.86 = 1.88$

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

First, calculate the sum of squared deviations from the mean. Let $x_i$ be the data - points and $\bar{x}$ be the mean.
\[

$$\begin{align*} \sum_{i = 1}^{9}(x_i-\bar{x})^2&=(1.86 - 2.743)^2+(2.28 - 2.743)^2+(2.44 - 2.743)^2+(2.9 - 2.743)^2+(3.02 - 2.743)^2+(3.05 - 2.743)^2+(3.65 - 2.743)^2+(3.7 - 2.743)^2+(3.74 - 2.743)^2\\ \end{align*}$$

\]
\[

$$\begin{align*} s&=\sqrt{\frac{\sum_{i = 1}^{9}(x_i-\bar{x})^2}{n - 1}}\\ \end{align*}$$

\]
\[

$$\begin{align*} s&\approx0.7095 \end{align*}$$

\]

Step8: Calculate the inter - quartile range (IQR)

$IQR=Q_3 - Q_1=3.675−2.36 = 1.315$

Step9: Calculate the lower and upper fences

Lower fence $=Q_1-1.5\times IQR=2.36-1.5\times1.315=2.36 - 1.9725 = 0.3875$
Upper fence $=Q_3+1.5\times IQR=3.675 + 1.5\times1.315=3.675+1.9725 = 5.6475$

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

$s^2=(0.7095)^2 = 0.5034$

Answer:

(a) $2.36$
(b) $3.675$
(c) $2.743$
(d) $3.02$
(e) $1.88$
(f) $0.7095$
(g) $1.315$
(h) Lower: $0.3875$, Upper: $5.6475$
(i) $0.5034$