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
