Write the expression: the sum of the quantity h and 3 divided by 6.

a) h/6 + 3
b) h/3 + 6
c) h+3/6
d) 6/h+3

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