In: Statistics and Probability
4)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 20substandard welds?
A)0.2033
B)0.6377
C)0.8132
D)0.4066
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)= 3.7749
P ( 10 ≤ X ≤ 20
)
Z1 = (X1 - µ ) / σ = -1.32
Z2 = (X2- µ ) / σ = 1.32
P ( -1.32 ≤ Z ≤ 1.32 )
= P ( Z ≤ 1.325 ) - P ( Z ≤
-1.325 ) = 0.8132 (answer)