Answer to Question #141133 in Statistics and Probability for Vaibhav lipare

Question

on an average a box containing 10 articles is likely to have 2 defectives. if we consider the consignment of 100 boxes, how many of them are expected to three or less defectives ?

Answer 1

Binomial distribution

$P=\frac{2}{10}=0.2, n=10$since average $np=2$ for 10 articles.

The probability of a box having three or less defectives is$P(\le3)$

$P(\le3)=P(0)+P(1)+P(2)+P(3)$

$=\sum_{x=0}^3\binom{10}{x}P^x(1-P)^{10-x}, x=0,1,2...10$

=0.879

Thus, in a consignment of 100 boxes, $0.879×100=87.9$ which is approximately 88 boxes are expected to have three or less defective articles.