In: Statistics and Probability
3.5.18. A shirt manufacturer knows that, on the average, 2% of his product will not meet quality specifications. Find the greatest number of shirts constituting a lot that will have, with probability 0.95, fewer than five defectives.
Question is from a chapter on Central Limit Theorem and Chebyshev’s theorem.
Here if the number of shirts are = n
Pr(Defective) = 0.02
so here expective number of deffective shirts = 0.02n
standard deviation of number of defective shirts = (0.02 * 0.98 * n)
so here, by using the central limit theorem
Pr(X < 5 ; 0.02n; 0.14 n] = 0.95
Z value = 1.645
(5 - 0.02n)/ (0.14n) = 1.645
5 - 0.02n = 0.23 sqrt (n)
5 - 0.02n = 0.23
25 + 0.0004n2 - 0.2529n = 0
so here by solving it we get
n = 122.27 or 123