Answer:
a) Y 0 1 2
P(Y) 0.58 0.23 0.11
b) mean= 0.45, S.D= 0.6718
c) mean= 1.285, S.D= 8.74
Step-by-step explanation:
a) The following table shows the probability distribution of X:
X 0 1 2 3 4 or more
P(X) 0.58 0.23 0.11 0.05 0.03
Defect >2 = cannot be sold
Y = the number of defects on a fat quarter that can be sold by Company F.
Y = defect that can be sold
Y = Defect less or equal to 2 = 0,1,2
Probability distribution of the random variable Y:
Y 0 1 2
P(Y) 0.58 0.23 0.11
b) mean of Y (μ)
μ = Σ x*P(Y)
= (0*0.58) +(1*0.23)+(2*0.11)
= 0+0.23+0.22 = 0.45
Standard deviation of Y = σ
σ = Σ√(x-mean)^2*P(Y)
= Σ√[(x- μ )^2*P(Y)]
= √[(0-0.45)^2*0.58+ (1-0.45)^2*0.23 + (2-0.45)^2*0.11]
= √[0.11745 + 0.069575 +0.264275
= √(0.4513
σ = 0.6718
Company G:
σ for defect that be sold = 0.66
μ for defect that be sold = 0.40
Difference between μ of F and μ of G
= 0.45-0.40 = 0.05
Difference between σ of F and σ of G
= 0.67-0.66 = 0.01
Selling price of fat quarter without defect = $5
Discount per defect = $1.5
Selling price per defect = 5-1.5 = $3.5
Discount per 2 defect = $1.5*2 = $3
Selling price per defect = 5-3 = $2
Since defect to be sold cannot be greater than 2, let Y = 5,3,2
Probability distribution of the selling price Y:
Y 5 3 2
P(Y) 0.58 0.23 0.11
μ = (5*0.58) +(3.5*0.23)+(2*0.11)
μ = 2.9+0.805+0.22 =1.285
σ = Σ√[(x- μ )^2*P(Y)]
σ = √[(5-1.285)^2*0.58+ (3-1.285)^2*0.23 + (2-1.285)^2*0.11]
σ = 8.00+0.68+0.06 = 8.74
= 10