91影视

Let X is the number of observed failures. The probability of failure is p.

It is given that

$p=0.10$and $n=20$

Consider,

$$\begin{gathered} P\left(x鈮�1\right)=P\left(x=0\right)+P\left(x=1\right) \\ =\left[\left(\begin{array}{c}20\\ 0\end{array}\right){\left(0.10\right)}^0{\left(0.90\right)}^{20-0}\right]+\left[\left(\begin{array}{c}20\\ 1\end{array}\right){\left(0.10\right)}^1{\left(0.90\right)}^{20-1}\right] \\ =\left(1脳1脳0.1216\right)+\left(20脳0.10脳0.{90}^{19}\right) \\ =\left(0.1216\right)+\left(0.2702\right) \\ =0.3918 \\ 鈮�0.392 \end{gathered}$$

Thus, the probability of$P\left(x鈮�1\right)$ is 0.392.

b.

By the reliability analysis, the level of the confidence for concluding that the true failure rate is less than or equal to p is defined as follows:

Level of confidence$\mathbf{=}\mathbf{1}\mathbf{-}\mathit{P}\left(x鈮�K\right)$

From the part a, the level of confidence is as follows:

Level of confidence

$$\begin{gathered} =1-P\left(x鈮�K\right) \\ =1-P\left(x鈮�K\right) \\ =1-0.392 \\ =0.608 \end{gathered}$$

Thus, the level of confidence for the part a. is 0.608.

Here, the level of confidence is not close to the probability value 1. This clearly indicates that this cannot be an acceptable level.

c.

From the information, it is clear that the sample size is n=20.

Increase the sample size n for 20 to 25 with

Consider,

$$\begin{gathered} P\left(x鈮�1\right)=P\left(x=0\right)+P\left(x=1\right) \\ =\left[\left(\begin{array}{c}25\\ 0\end{array}\right){\left(0.10\right)}^0{\left(0.90\right)}^{25-0}\right]+\left[\left(\begin{array}{c}25\\ 1\end{array}\right){\left(0.10\right)}^1{\left(0.90\right)}^{25-1}\right] \\ =\left(1脳10.0718\right)+\left(20脳0.10脳0.{90}^{24}\right) \\ =\left(0.0718\right)+\left(0.1994\right) \\ =0.2712 \\ 鈮�0.271 \end{gathered}$$

Thus, the probability of $P\left(x鈮�1\right)$is 0.271.

The level of the confidence with$\mathit{p}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}$and$\mathit{n}\mathbf{=}\mathbf{25}$can be obtained as follows:

Level of confidence is:

$$\begin{gathered} 1-P\left(x鈮�K\right)=1-P\left(x鈮�K\right) \\ =1-0.271 \\ =0.729 \end{gathered}$$

Thus, the level of confidence is 0.729.

From the information, it is clear that the sample size is n=20

Decrease the number of failures 1 to 0.

Consider,

$$\begin{gathered} P\left(x鈮�0\right)=P\left(x=0\right) \\ =\left(\begin{array}{c}25\\ 0\end{array}\right){\left(0.10\right)}^0{\left(0.90\right)}^{20-0} \\ =0.1216 \\ 鈮�0.122 \end{gathered}$$

Thus, the probability of$P\left(x鈮�0\right)$is 0.122.

The level of confidence with$\mathit{p}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}$ and$\mathit{n}\mathbf{=}\mathbf{20}$ can be obtained as follows:

Level of confidence$\mathbf{=}\mathbf{1}\mathbf{-}\mathit{P}\left(x鈮�K\right)$

That is

$$\begin{gathered} 1-P\left(X鈮�K\right)=1-P\left(x鈮�0\right) \\ =1-0.122 \\ =0.878 \end{gathered}$$

Thus, the level of confidence is 0.878.

Here, the level of confidence by increasing the sample size from 20 to 25 is 0.729 whereas by decreasing the number of failure 1 to 0 is 0.878.

Hence, it is clear that the confidence level is increased by decreasing the number of K failures that has been allowed in the sample.

d.

Consider,

K=0,

It is necessary to obtain the value of n. Thus, the level of confidence should be at least 0.95. That is,

$$\begin{gathered} P\left(x=0\right)鈮�0.95 \\ P\left(x=0\right)=\left\{\left(\begin{array}{c}n\\ 0\end{array}\right){\left(p\right)}^x{\left(q\right)}^{n-x}\right\}鈮�0.05 \\ =\left\{\frac{n!}{0!n!}{\left(0.10\right)}^0{\left(0.90\right)}^{n-0}\right\}鈮�0.05 \\ =\left\{1脳1脳{\left(0.90\right)}^n\right\}鈮�0.05 \\ =\left\{{\left(0.90\right)}^n\right\}鈮�0.05 \end{gathered}$$

Taking logarithm of both sides,

$$\begin{gathered} \mathrm{In}\left[{\left(0.90\right)}^{\mathrm{n}}\right]鈮�\mathrm{In}\left[0.05\right] \\ \mathrm{n}\mathrm{In}\left(0.90\right)鈮�\mathrm{In}\left(0.05\right) \\ \mathrm{n}鈮�\frac{\mathrm{In}\left(0.05\right)}{\mathrm{In}\left(0.90\right)} \\ \mathrm{n}鈮�\frac{-2.9957}{-0.1054} \\ \mathrm{n}鈮�28.4 \\ \mathrm{n}鈮�29 \end{gathered}$$

Thus, the value of n when $K=0$is 29 in order to get the level of confidence at least 0.95.

Consider, K=1.

It is necessary to obtain the value of n. Thus, the level of confidence should be at least 0.95. That is,$P(x=0)鈮�0.95$

role="math" localid="1660716120882" $$\begin{gathered} P\left(x鈮�1\right)=\left\{\left(\begin{array}{c}n\\ 0\end{array}\right){\left(p\right)}^x{\left(q\right)}^{n-0}+\left(\begin{array}{c}n\\ 0\end{array}\right){\left(p\right)}^x{\left(q\right)}^{n-1}\right\}鈮�0.05 \\ =\left\{\frac{n!}{0!n!}{\left(0.10\right)}^0{\left(0.90\right)}^{n-0}+\frac{n!}{1!\left(n-1\right)!}{\left(0.10\right)}^1{\left(0.90\right)}^{n-1}\right\}鈮�0.05 \\ =\left\{1脳1脳{\left(0.90\right)}^n+n{\left(0.10\right)}^1{\left(0.90\right)}^{n-1}\right\}鈮�0.05 \\ ={\left(0.90\right)}^{n-1}\left\{0.90+n\left(0.10\right)\right\}鈮�0.05鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�..\left(1\right) \end{gathered}$$

By the trail, and error,

Consider, n=30 and substitute in the equation (1)

$$\begin{gathered} {\left(0.90\right)}^{\mathrm{n}-1}\left\{0.90+\mathrm{n}\left(0.10\right)\right\}={\left(0.90\right)}^{30-1}\left\{0.90+30\left(0.10\right)\right\} \\ =0.{90}^{29}\left(0.90+3\right) \\ =0.{90}^{29}\left(3.90\right) \\ =0.1837 \end{gathered}$$

Consider, n=40and substitute in the equation (1),

$$\begin{gathered} {\left(0.90\right)}^{\mathrm{n}-1}\left\{0.90+\mathrm{n}\left(0.10\right)\right\}={\left(0.90\right)}^{40-1}\left\{0.90+40(0.10\right\} \\ =0.{90}^{39}\left(0.90+4\right) \\ =0.{90}^{39}\left(4.90\right) \\ =0.0805 \end{gathered}$$

Consider,n=45 and substitute in the equation (1)

$$\begin{gathered} {\left(0.90\right)}^{\mathrm{n}-1}\left\{0.90+\mathrm{n}\left(0.10\right)\right\}={\left(0.90\right)}^{45-1}\left\{0.90+45(0.10\right\} \\ =0.{90}^{44}\left(0.90+4.5\right) \\ =0.{90}^{44}\left(5.4\right) \\ =0.0524 \end{gathered}$$

Consider, and substitute in the equation (1)

$$\begin{gathered} {\left(0.90\right)}^{\mathrm{n}-1}\left\{0.90+\mathrm{n}\left(0.10\right)\right\}={\left(0.90\right)}^{46-1}\left\{0.90+46(0.10\right\} \\ =0.{90}^{45}\left(0.90+4.6\right) \\ =0.{90}^{45}\left(5.5\right) \\ =0.0480 \end{gathered}$$

Thus, the value of n when$K=1$ is 46 in order to get the level of confidence at least 0.95.