Question

please show work.

6) 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 between 10 and 20 substandard welds?

Answer 1

0.05*N=0.05*7500=375

n<0.05*N

also, np = 300*0.05=15

n(1-p) = 285

np and n(1-p) ≥10 , so, binomial can be approximated to normal distribution

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Sample size , n =    300                              
Probability of an event of interest, p =   0.05                              
we need to calculate probability for ,                                  
10   < X <   20                          
Mean = np =    15                              
std dev ,σ=√np(1-p)=   3.775                              
                                  
P (   10   < X <   20   ) = P (   10.5   ≤ Xnormal ≤   19.5      
Z1 =   (X1normal - µ ) / σ =   -1.192                          
Z2 =   (X2 normal- µ ) / σ =   1.192                          
                                  
P (    -1.192   < Z <   1.192   )                   
                                  
= P ( Z <   1.192   ) - P ( Z <   -1.192   ) =    0.8834   -    0.1166   =    0.7668

please revert for doubt...