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

0.0307

Step-by-step explanation:

Given data;

Rivets from vendor A = 70%;     = 0.7

Rivets from vendor B = 80%;     = 0.8

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

Assuming that and represents the mean numbers of rivets that hits the targeted specification.

Using the Central Limit Theorem to evaluate the Binomial distribution of   and ; Then:

= 500 × 0.7

       = 350

       > 10

= 500 × 0.3

       = 150

       > 10

 =  500 × 0.8

       = 400

       > 10

=  500 × 0.2

       = 100

       > 10

Hence;

≅ N ( , ) = N (350,105)

 ≅ N (  , ) = N (400, 80)

T =  +

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:

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

529 and 531

step-by-step explanation: