Question

Find the mean and compare it with the median. Find the standard deviation and compare it with the interquartile range

Answer 1

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)

`Mean = 4`

`Median=3`

`Mean > Median`

(b)

`IQR = 2`

`SD = 3.2`

`SD> IQR`

Explanation:

Given:

Data: 4, 1, 3, 10, 2.

Solving (a): The mean and the Median

The mean is calculated as follows:

`Mean = (1)/(n) \sum x`

Where

`n = 5` i.e 5 data

So, the expression becomes:

`Mean = (1)/(5)(4 + 1+3+10+2)`

`Mean = (1)/(5)(20)`

`Mean = 4`

Calculating the Median:

First, arrange the order (ascending order):

`Data: 1, 2,3,4,10`

Because n is odd

The median is represented as:

`Median = ((n+1)/(2))th\ item`

`Median = ((5+1)/(2))th\ item\\`

`Median = ((6)/(2))th\ item`

`Median = 3rd\ item`

From the arranged data, the 3rd item is 3.

Hence:

`Median=3`

By comparison, the mean is greater than the median because
`4 > 3`

Solving (b): Standard Deviation and IQR

The standard deviation is calculated as follows:

`SD= \sqrt{(\sum (x_i - Mean)^2)/(n)`

So, we have:

`SD= \sqrt{((4 - 4)^2+(1 - 4)^2+(3 - 4)^2+(10 - 4)^2+(2 - 4)^2)/(5)`

`SD= \sqrt{((0)^2+(- 3)^2+(- 1)^2+(6)^2+(-2)^2)/(5)`

`SD= \sqrt{(0+9+1+36+4)/(5)`

`SD= \sqrt{(50)/(5)`

`SD= √(10)`

`SD = 3.2`

Calculating IQR

`IQR = Q_3 - Q_1`

`Data: 1, 2,3,4,10`

In (a), we calculate the median as:

`Median=3`

First, we calculate
`Q_1`

`Q_1 = Median\ of\ the\ first\ half.`

The first half is:

`First\ Half = 1,2,3`

So:

`Q_1 = 2`

First, we calculate
`Q_3`

`Q_3 = Median\ of\ the\ second\ half.`

`Second\ Half = 3,4,10`

So:

`Q_3 = 4`

Recall that:

`IQR = Q_3 - Q_1`

`IQR = 4 - 2`

`IQR = 2`

So, we have:

`IQR = 2`

`SD = 3.2`

By comparison, the standard deviation is greater than the IQR because:

`3.2 > 2`

So,

`SD> IQR`