Smith is a weld inspector at a shipyard. He knows from keeping track of good and...

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?

Solutions

Expert Solution

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

orchestra answered 2 years ago

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