EX 3.31 binding strength test

• Given:
  • $20\%$ of all copies fail the binding strength test
• Definition:
  • Let $X$ denote the number of failures among randomly selected copies
• Distribution:
  • $X$ follows a binomial distribution
  • Parameters:
    • $n=15$ (number of selected copies)
    • $p=0.2$ (probability of failure)

1. The probability that at most 8 fail the test is $P\left(X\le 8\right)=\sum _{y=0}^{8}b\left(y;15,.2\right)=B\left(8;15,.2\right)$ which is the entry in the $x=8$ row and the $p=.2$ column of the $n=15$ binomial table. From Appendix Table A.1, the probability is $B\left(8;15,.2\right)=.999$ .

2. The probability that exactly 8 fail is $P\left(X=8\right)=P\left(X\le 8\right)-P\left(X\le 7\right)=B\left(8;15,.2\right)-B\left(7;15,.2\right)$ which is the difference between two consecutive entries in the $p=.2$ column. The result is $.999-.996=.003$ .

3. The probability that at least 8 fail is

$$\begin{array}{rl}P(X\ge 8)& =1-P(X\le 7)\\ & =1-B(7;15,0.2)\\ & =1-\left(\begin{array}{l}\text{entry in }x=7\\ \text{row of }p=0.2\text{ column}\end{array}\right)\\ & =1-0.996\\ & =0.004\end{array}$$

4. Finally, the probability that between 4 and 7, inclusive, fail is

$$\begin{array}{rl}P(4\le X\le 7)& =P(X=4,5,6,\text{ or }7)\\ & =P(X\le 7)-P(X\le 3)\\ & =B(7;15,0.2)-B(3;15,0.2)\\ & =0.996-0.648\\ & =0.348\end{array}$$

Notice that this latter probability is the difference between entries in the $x=7$ and $x=3$ rows, not the $x=7$ and $x=4$ rows.