Question

Smith is a weld inspector at a shipyard. He knows from keeping track of good and substandard welds that for the afternoon shift 5% of all welds done will be substandard. If Smith checks 300 welds completed that shift, what is the probability that he will find more than 25 substandard welds?

Answer 1

Solution:

Given that,

P = 0.05

1 - P = 0.95

n = 300

Here, BIN ( n , P ) that is , BIN (300 , 0.05)

then,

n*p = 300*0.05 = 15 > 5

n(1- P) = 300*0.95 = 285 > 5

According to normal approximation binomial,

X Normal

Mean = = n*P = 15

Standard deviation = =n*p*(1-p) = 300*0.05*0.95 = 14.25

We using countinuity correction factor

P(x > a ) = P( X > a + 0.5)

P(x > 25.5) = 1 - P(x < 25.5)

= 1 - P((x - ) / < (25.5 - 15) / 14.25)

= 1 - P(z < 2.78)

= 1 - 0.9973   

= 0.0027

Probability = 0.0027

The probability that he will find more than 25 substandard welds is 0.0027