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)

$IQR = 2$

$SD = 3.2$

$SD> IQR$

Step-by-step explanation:

Given:

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

Solving (a): The mean and the Median

The mean is calculated as follows:

$Mean = \frac{1}{n} \sum x$

Where

$n = 5$ i.e 5 data

So, the expression becomes:

$Mean = \frac{1}{5}(4 + 1+3+10+2)$

$Mean = \frac{1}{5}(20)$

$Mean = 4$

Calculating the Median:

First, arrange the order (ascending order):

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

Because n is odd

The median is represented as:

$Median = (\frac{n+1}{2})th\ item$

$Median = (\frac{5+1}{2})th\ item$

$Median = (\frac{6}{2})th\ item$

$Median = 3rd\ item$

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:

$SD= \sqrt{\frac{\sum (x_i - Mean)^2}{n}$

So, we have:

$SD= \sqrt{\frac{(4 - 4)^2+(1 - 4)^2+(3 - 4)^2+(10 - 4)^2+(2 - 4)^2}{5}$

$SD= \sqrt{\frac{(0)^2+(- 3)^2+(- 1)^2+(6)^2+(-2)^2}{5}$

$SD= \sqrt{\frac{0+9+1+36+4}{5}$

$SD= \sqrt{\frac{50}{5}$

$SD= \sqrt{10}$

$SD = 3.2$

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,

$SD> IQR$