Question:
Find the mean and compare it with the median. Find the standard deviation and compare it with the interquartile range. Calculate s for the data 4, 1, 3, 10, 2.
Answer:
(a)
[tex]Mean = 4[/tex]
[tex]Median=3[/tex]
[tex]Mean > Median[/tex]
(b)
[tex]IQR = 2[/tex]
[tex]SD = 3.2[/tex]
[tex]SD> IQR[/tex]
Step-by-step explanation:
Given:
Data: 4, 1, 3, 10, 2.
Solving (a): The mean and the Median
The mean is calculated as follows:
[tex]Mean = \frac{1}{n} \sum x[/tex]
Where
[tex]n = 5[/tex] i.e 5 data
So, the expression becomes:
[tex]Mean = \frac{1}{5}(4 + 1+3+10+2)[/tex]
[tex]Mean = \frac{1}{5}(20)[/tex]
[tex]Mean = 4[/tex]
Calculating the Median:
First, arrange the order (ascending order):
[tex]Data: 1, 2,3,4,10[/tex]
Because n is odd
The median is represented as:
[tex]Median = (\frac{n+1}{2})th\ item[/tex]
[tex]Median = (\frac{5+1}{2})th\ item\\[/tex]
[tex]Median = (\frac{6}{2})th\ item[/tex]
[tex]Median = 3rd\ item[/tex]
From the arranged data, the 3rd item is 3.
Hence:
[tex]Median=3[/tex]
By comparison, the mean is greater than the median because [tex]4 > 3[/tex]
Solving (b): Standard Deviation and IQR
The standard deviation is calculated as follows:
[tex]SD= \sqrt{\frac{\sum (x_i - Mean)^2}{n}[/tex]
So, we have:
[tex]SD= \sqrt{\frac{(4 - 4)^2+(1 - 4)^2+(3 - 4)^2+(10 - 4)^2+(2 - 4)^2}{5}[/tex]
[tex]SD= \sqrt{\frac{(0)^2+(- 3)^2+(- 1)^2+(6)^2+(-2)^2}{5}[/tex]
[tex]SD= \sqrt{\frac{0+9+1+36+4}{5}[/tex]
[tex]SD= \sqrt{\frac{50}{5}[/tex]
[tex]SD= \sqrt{10}[/tex]
[tex]SD = 3.2[/tex]
Calculating IQR
[tex]IQR = Q_3 - Q_1[/tex]
[tex]Data: 1, 2,3,4,10[/tex]
In (a), we calculate the median as:
[tex]Median=3[/tex]
First, we calculate [tex]Q_1[/tex]
[tex]Q_1 = Median\ of\ the\ first\ half.[/tex]
The first half is:
[tex]First\ Half = 1,2,3[/tex]
So:
[tex]Q_1 = 2[/tex]
First, we calculate [tex]Q_3[/tex]
[tex]Q_3 = Median\ of\ the\ second\ half.[/tex]
[tex]Second\ Half = 3,4,10[/tex]
So:
[tex]Q_3 = 4[/tex]
Recall that:
[tex]IQR = Q_3 - Q_1[/tex]
[tex]IQR = 4 - 2[/tex]
[tex]IQR = 2[/tex]
So, we have:
[tex]IQR = 2[/tex]
[tex]SD = 3.2[/tex]
By comparison, the standard deviation is greater than the IQR because:
[tex]3.2 > 2[/tex]
So,
[tex]SD> IQR[/tex]
