Question

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.

Answer 1

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