Question 1168022
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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 of the 7500 welds completed that shift, what is the probability 
that he will find less than 20 substandard welds?
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        In this problem,  the info about  " 7500 welds completed that shift "  is excessive.

        It does not add any sense to the problem, does not produce any influence on the solution 

        and is not used in the solution.



<pre>
So, it is a binomial experiment with the number of trials n= 300.
The "success" is detecting a substandard weld.  
The probability of the individual success is p= 0.05.
The problem asks about the probability to have 19 or less individual successes.


Since the number of trials is great (more than 25), it is recommended to use
a normal approximation to binomial distribution.


For this normal distribution, the mean is  p*n = 0.05*300 = 15,

and the standard deviation is S = {{{sqrt(p*n*(1-p))}}} = {{{sqrt(0.05*300*(1-0.05))}}} = 3.7749 (rounded).


Now you can use standard function normalcdf(z1, z2, m, S) in your regular calculator TI-83/84


    p = normalcdf(-9999, 19.5, 15, 3.7749) = 0.8834.    <U>ANSWER</U>


Here 19.5  is the continuity correction factor.


Alternatively, you may use an online calculator at web-site

    https://onlinestatbook.com/2/calculators/normal_dist.html

which has a convenient interface and provides a visual picture of the area of interest under the normal curve.
</pre>

Solved.