Answer to Question #174570 in Statistics and Probability for Kim smith

Question

a toy manufacturer uses this acceptance sampling plan: Randomly select and test 47 batteries and determine what batteries do not meet specifications. A shipment contains 4000 batteries, and 1% of them do not meet specifications. What is the prob be accepted , or will many be rejected?

Answer 1

Solution:

Let $X$ be the number of defective batteries.

The probability distribution of the random variable $X$, hypergeometric distribution, is

$h(x ; n, M, N)=\frac{\left(\begin{array}{l} M \\ x \end{array}\right)\left(\begin{array}{l} N-M \\ n-x \end{array}\right)}{\left(\begin{array}{l} N \\ n \end{array}\right)}$

Given: $N=4000, M=4000 \times 0.01=40, n=47$ .

This part is missing, so we assume we are asked $P(X\le 3)$.

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

$\begin{array}{c} =\frac{\left(\begin{array}{c} 40 \\ 0 \end{array}\right)\left(\begin{array}{c} 4000-40 \\ 47-0 \end{array}\right)}{\left(\begin{array}{c} 4000 \\ 47 \end{array}\right)}+\frac{\left(\begin{array}{c} 40 \\ 1 \end{array}\right)\left(\begin{array}{c} 4000-40 \\ 47-1 \end{array}\right)}{\left(\begin{array}{c} 4000 \\ 47 \end{array}\right)}+ \\ +\frac{\left(\begin{array}{c} 40 \\ 2 \end{array}\right)\left(\begin{array}{c} 4000-40 \\ 47-2 \end{array}\right)}{\left(\begin{array}{c} 4000 \\ 47 \end{array}\right)}+\frac{\left(\begin{array}{c} 40 \\ 3 \end{array}\right)\left(\begin{array}{c} 4000-40 \\ 47-3 \end{array}\right)}{\left(\begin{array}{c} 4000 \\ 47 \end{array}\right)} \end{array}$

$\approx 0.62181+0.29867+0.06843+0.00996 \\ \approx 0.99887$