Step-by-step explanation:
Given:-
- The Values ( Attach by the User )
To find:-
Solution:- (Q.1)
[tex]\sf \: Arithmetic \: mean = \dfrac{Sum \: of \: all \: observation}{No.of \: observations}\\\\\\: \implies \sf \: mean \: = \dfrac{1 + 1 + 1 + 1 + 3 + 3 + 4 + 4 + 4 + 8 + 12}{11}\\\\\\ : \implies \sf \: mean = \dfrac{42}{11} \\\\\\ : \implies \sf \: mean = 3.81[/tex]
Solution:- (Q.2)
[tex]: \implies \sf \:median = \bigg({\dfrac{n + 1}{2} \bigg)^{th}}\\\\\\ :\implies \sf \:median = \bigg( {\dfrac{11 + 1}{2} \bigg)^{th}}\\\\\\: \implies \sf \:median = \bigg( \dfrac{12}{2} \bigg) ^{th}\\\\\\ : \implies \sf \:median = 6th \: item\\\\\\ : \implies \sf \:median = 3[/tex]
Solution:- (Q.3)
Mode = Number of Maximum times occurring in Observation.
Mode = 1 Because its Occurring most in the Observation.
Solution:- (Q.4):-
Range = Highest - Lowest
⠀⠀⠀⠀= ⠀⠀12 ⠀⠀- ⠀1
⠀⠀⠀⠀= ⠀⠀⠀⠀⠀11
.°. 11 is the Required Range
