Question

Smith is a weld inspector at a shipyard. He knows from keeping track of good and substandard welds that for the afternoon shift 4.5% of all welds done will be substandard. If Smith checks 250 of the 7500 welds completed in an afternoon shift, then what is the probability that he will find less than 15 substandard welds?

Answer 1

Let p be the proportion of the substandard welds and x be the number of subtandard welds.

Given : p = 0.045 , n = 250

$1596248464420_blob.png$ = x /n = 15/250 = 0.06

According to sampling distribution of sample proportion $1596248520224_blob.png$ , The sample proportion approximately follows normal distribution with mean $1596248520244_blob.png$ = P  and standard deviation $1596248520224_blob.png$ =   

P*(1-P) n

Therefore $1596248520244_blob.png$ = 0.045

and $1596248520224_blob.png$ = $\sqrt{\frac{0.045*(1-0.045)}{250}}$ = 0.0131

We are asked to find P( $1596248464420_blob.png$ < 0.06 )

= $P\left ( \frac{\hat{p}-\mu _{\hat{p}}}{\sigma _{\hat{p}}} \leq\frac{0.06-0.045}{0.0131}\right )$

= P( z ≤ 1.14 )

= 0.8729

The probability that he will find less than 15 substandard welds is 0.8729