Question

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

Answer 1

Answer:

0.0307

Step-by-step explanation:

Given data;

Rivets from vendor A = 70%;     $P__A$ = 0.7

Rivets from vendor B = 80%;     $P__B$ = 0.8

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

Assuming that $Y__A}$ and $Y__B$ represents the mean numbers of rivets that hits the targeted specification.

Using the Central Limit Theorem to evaluate the Binomial distribution of  $Y__A}$ and $Y__B$; Then:

$nP__A$ = 500 × 0.7

       = 350

       > 10

$nP__A}(1-P_A)$ = 500 × 0.3

       = 150

       > 10

$nP__B$  =  500 × 0.8

       = 400

       > 10

$nP__B}(1-P_B)$ =  500 × 0.2

       = 100

       > 10

Hence;

$Y__A$ ≅ N ( $nP__A$ , $nP__A}(1-P_A)$ ) = N (350,105)

$Y__B$  ≅ N ( $nP__B$  , $nP__B}(1-P_B)$ ) = N (400, 80)

T = $Y__A$  + $Y__B$

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:

$z= \frac{775.5 -750}{\sqrt{185} }$

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