Updated Solution of STA 301 Assignment 1 of 2018

Q-1.

If for any moderately asymmetrical distribution mode =15, and mean= 10.5 then find median?

Answer 1:             

Mode =3Median-2 Mean     
            3Median=Mode+2Mean   
            3Median=15+2(10.5)        
            3Median =15+21              
            3Median=36
            Median=36/3
            Median =12  

Answer 2:             

Height	Frequency	Class Boundaries	Cumulative Frequency
60-62	5	59.5-62.5	5
63-65	18	62.5-65.5	23
66-68	42	65.5-68.5	65
69-71	27	68.5-71.5	92
72-74	8	71.5-74.5	100
Total	100

median = l+h/f (n/2-c)

=n/2 =100/2=50         

median=  65.5+3/42(50-23)   

median= 67.43          

    H.M           

Height	Frequency	X	F*(1/X)
60-62	5	61	0.08
63-65	18	64	0.28
66-68	42	67	0.63
79-71	27	70	0.39
72-74	8	73	0.11
Total	100		1.49

    H.M = n/ Σ f(1/x)   

    H.M=  100/1.49     

    H.M= 67.11