Question

Item 4 Find the median, first quartile, third quartile, and interquartile range of the data. 40,33,37,54,41,34,27,39,35 Item 4 Find the median, first quartile, third quartile, and interquartile range of the data. 40,33,37,54,41,34,27,39,35

Answer 1

Answer:

Median =37

First quartile = 33.5

Third quartile=40.5

Interquartile=7

Step-by-step explanation:

Given data

40,33,37,54,41,34,27,39,35

First we have to rearrange the data in either ascending order or descending order.

We can rewrite the data as

27,33,34,35,37,39,40,41,54

Median:

The middle term of the data is the median of the data.

The number of data = n

If n is odd, the median of the data= $(\frac{ n+1}2)^{th$ term

If n is even, the median of the data $\frac{(\frac {n}{2})^{th} +{(\frac {n}{2}+1)^{th}}}{2}$

Here n= 9

The median of the data= $(\frac{9+1}{2})^{th}$ term

                                       $=(\frac{10}{2})^{th}$ term

                                        $=5^{th}$ term

                                       =37

First quartile:

The median term of the lower half of the data.

Lower half = 27,33,34,35

The median of the lower half = $\frac{(\frac 42)^{th} term+(\frac 42+1)^{th}term}{2}$

                                               = $\frac{(2)^{nd} term+(3)^{rd}term}{2}$

                                              = $\frac{33+34}2$

                                               =33.5

The first quartile= 33.5

Third  quartile:

The median term of the upper half of the data.

Upper half 39,40,41,54

The median of the lower half = $\frac{(\frac 42)^{th} term+(\frac 42+1)^{th}term}{2}$

                                                = $\frac{(2)^{nd} term+(3)^{rd}term}{2}$

                                                = $\frac{40+41}2$

                                                 =40.5

Interquartile:

The difference between first quartile and third quartile.

Interquartile = 40.5-33.5

                       =7