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damaskus [11]
2 years ago
11

Two different suppliers, A and B, provide a manufacturer with the same part. All supplies of this part are kept in a large bin.

in the past, 5% of the parts supplied by A and 9% of the parts supplied by B have been defective. A supplies four times as many parts as B. Suppose you reach into the bin and select a part, and find it is nondetective. What is the probability that it was supplied by A
Mathematics
1 answer:
koban [17]2 years ago
5 0

Answer:

The probability of selecting a non-defective part provided by supplier A is 0.807.

Step-by-step explanation:

Let <em>A</em> = a part is supplied by supplier A, <em>B</em> = a part is supplied by supplier B and <em>D</em> = a part is defective.

<u>Given</u>:

P (D|A) = 0.05, P(D|B) = 0.09

A supplies four times as many parts as B, i.e. n (A) = 4 and n (B) = 1.

Then the probability of event <em>A</em> and <em>B</em> is:

P(A)=\frac{n(A)}{n(A)+N(B)}= \frac{4}{4+1}=0.80\\P(B)\frac{n(B)}{n(A)+N(B)}= \frac{1}{4+1}=0.20

Compute the probability of selecting a defective product:

P(D)=P(D|A)P(A)+P(D|B)P(B)\\=(0.05\times0.80)+(0.09\times0.20)\\=0.058

The probability of selecting a non-defective part provided by supplier A is:

P(A|D')=\frac{P(D'|A)P(A)}{P(D')} = \frac{(1-P(D|A))P(A)}{1-P(D)}\\=\frac{(1-0.05)\times0.80}{(1-0.058)}\\ =0.80679\\\approx0.807

Thus, the probability of selecting a non-defective part provided by supplier A is 0.807.

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Company F sells fabrics known as fat quarters, which are rectangles of fabric created by cutting a yard of fabric into four piec
jeka94

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

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