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Seventy percent of rivets from vendor A meet a certain strength specification, and 80% of rivets from vendor B meet the same specification. If 500 rivets are purchased from each vendor, what is the probability that more than 775 of the rivets meet the specifications

Respuesta :

ajeigbeibraheem

Answer:

0.0307

Step-by-step explanation:

Given data;

Rivets from vendor A = 70%;     [tex]P__A[/tex] = 0.7

Rivets from vendor B = 80%;     [tex]P__B[/tex] = 0.8

Number of rivets purchased from each vendor (n) =500

Assuming that [tex]Y__A}[/tex] and [tex]Y__B[/tex] represents the mean numbers of rivets that hits the targeted specification.

Using the Central Limit Theorem to evaluate the Binomial distribution of  [tex]Y__A}[/tex] and [tex]Y__B[/tex]; Then:

[tex]nP__A[/tex] = 500 × 0.7

       = 350

       > 10

[tex]nP__A}(1-P_A)[/tex] = 500 × 0.3

       = 150

       > 10

[tex]nP__B[/tex]  =  500 × 0.8

       = 400

       > 10

[tex]nP__B}(1-P_B)[/tex] =  500 × 0.2

       = 100

       > 10

Hence;

[tex]Y__A[/tex] ≅ N ( [tex]nP__A[/tex] , [tex]nP__A}(1-P_A)[/tex] ) = N (350,105)

[tex]Y__B[/tex]  ≅ N ( [tex]nP__B[/tex]  , [tex]nP__B}(1-P_B)[/tex] ) = N (400, 80)

T = [tex]Y__A[/tex]  + [tex]Y__B[/tex]

T ≅ N (750, 185)

We are tasked to find out that what is the probability that more than 775 of the rivets meet the specifications; So, we need to determine that P(T > 775); the continuity  correction point T = 775.5

The corresponding Z-value is as follows:

[tex]z= \frac{775.5 -750}{\sqrt{185} }[/tex]

Z= 1.87

P(T>775) =P(Z>1.87)

               = 1 - 0.9693

               = 0.0307

∴ the probability that more than 775 of the rivets meet the specifications = 0.0307

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