Probability & Statistics Q&A || Mean(μ), Standard Deviation(σ), Coefficient of Variation (CV).

1 a) The lives of two models of refrigerators in a recent survey are:

Life (No of Years):	0-2	2-4	4-6	6-8	8-10	10-12
Model x:	5	16	13	7	5	4
Model Y:	2	7	12	19	9	1

(i) What is the average life of each model of these refrigerators?
(ii) Which model has a greater uniformity?

∴ Note: Tablular contents may not fit on your screen properly, please swipe left to view all contents. 

Ans: let's demonstrate these information into the tabular form.

No. of Year	Mid Value	Model A			Model B
	(x)	f	fx	fx²	f	fx	fx²
0-2	1	5	5	5	2	2	2
2-4	3	16	48	144	7	21	63
4-6	5	13	65	325	12	60	300
6-8	7	7	49	343	19	133	931
8-10	9	5	45	405	9	81	729
10-12	11	4	44	484	1	11	121
		Σf=50	Σfx=256	Σfx²=1706	Σf=50	Σfx=308	Σfx²=2146

For Model A:
Mean(μ)

Mean(μ) = Σfx/n
Mean(μ) = 256/50
Mean(μ) = 5.12
  

Standard Deviation(σ)

SD(σ) = √Σfx²/n-(Σfx/n)²
SD(σ) = √1706/50-(256/50)²
SD(σ) = √34.12-(5.12)²
SD(σ) = √34.12-26.2144
SD(σ) = √7.9056
SD(σ) = 2.81
  

For Model B:
Mean(μ)

Mean(μ) = Σfx/n
Mean(μ) = 308/50
Mean(μ) = 6.16
  

Standard Deviation(σ)

SD(σ) = √Σfx²/n-(Σfx/n)²
SD(σ) = √2146/50-(308/50)²
SD(σ) = √42.92-(6.16)²
SD(σ) = √42.92-37.9456
SD(σ) = √4.9744
SD(σ) = 2.23
  

Average life of Model A is = 5.12 Years
Average life of Model B is = 6.16 Years
(i) Therefore Model B is the average life.

Now, Coefficient of Variation (CV):

  CV of Model A = Standard Deviation(σ)/Mean(μ)*100%
                = 2.81/5.12*100%
                = 54.88%
                
  CV of Model B = Standard Deviation(σ)/Mean(μ)*100%
                = 2.23/6.16*100%
                = 36.20%             
  

(ii) Since CV of A is > than CV of B, So Model B has more uniformity.