Respuesta :
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= [tex](\frac{ n+1}2)^{th[/tex] term
If n is even, the median of the data [tex]\frac{(\frac {n}{2})^{th} +{(\frac {n}{2}+1)^{th}}}{2}[/tex]
Here n= 9
The median of the data= [tex](\frac{9+1}{2})^{th}[/tex] term
[tex]=(\frac{10}{2})^{th}[/tex] term
[tex]=5^{th}[/tex] 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 = [tex]\frac{(\frac 42)^{th} term+(\frac 42+1)^{th}term}{2}[/tex]
= [tex]\frac{(2)^{nd} term+(3)^{rd}term}{2}[/tex]
= [tex]\frac{33+34}2[/tex]
=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 = [tex]\frac{(\frac 42)^{th} term+(\frac 42+1)^{th}term}{2}[/tex]
= [tex]\frac{(2)^{nd} term+(3)^{rd}term}{2}[/tex]
= [tex]\frac{40+41}2[/tex]
=40.5
Interquartile:
The difference between first quartile and third quartile.
Interquartile = 40.5-33.5
=7
