Answer:
Step-by-step explanation:
Given that:
population mean = 10
standard deviation = 0.1
sample mean = 9.8 < x > 10.2
The z score can be computed as:
![z = \dfrac{\bar x - \mu}{\sigma}](https://tex.z-dn.net/?f=z%20%3D%20%5Cdfrac%7B%5Cbar%20x%20-%20%5Cmu%7D%7B%5Csigma%7D)
if x > 10.2
![z = \dfrac{10.2- 10}{0.1}](https://tex.z-dn.net/?f=z%20%3D%20%5Cdfrac%7B10.2-%2010%7D%7B0.1%7D)
![z = \dfrac{0.2}{0.1}](https://tex.z-dn.net/?f=z%20%3D%20%5Cdfrac%7B0.2%7D%7B0.1%7D)
z = 2
If x < 9.8
![z = \dfrac{9.8- 10}{0.1}](https://tex.z-dn.net/?f=z%20%3D%20%5Cdfrac%7B9.8-%2010%7D%7B0.1%7D)
![z = \dfrac{-0.2}{0.1}](https://tex.z-dn.net/?f=z%20%3D%20%5Cdfrac%7B-0.2%7D%7B0.1%7D)
z = -2
The p-value = P (z ≤ 2) + P (z ≥ 2)
The p-value = P (z ≤ 2) + ( 1 - P (z ≥ 2)
p-value = 0.022750 +(1 - 0.97725)
p-value = 0.022750 + 0.022750
p-value = 0.0455
Therefore; the probability of defectives = 4.55%
the probability of acceptable = 1 - the probability of defectives
the probability of acceptable = 1 - 0.0455
the probability of acceptable = 0.9545
the probability of acceptable = 95.45%
4.55% are defective or 95.45% is acceptable.
sampling distribution of proportions:
sample size n=1000
p = 0.0455
The z - score for this distribution at most 5% of the items is;
![z = \dfrac{0.05 - 0.0455}{\sqrt{\dfrac{0.0455\times 0.9545}{1000}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cdfrac%7B0.05%20-%200.0455%7D%7B%5Csqrt%7B%5Cdfrac%7B0.0455%5Ctimes%200.9545%7D%7B1000%7D%7D%7D)
![z = \dfrac{0.0045}{\sqrt{\dfrac{0.04342975}{1000}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cdfrac%7B0.0045%7D%7B%5Csqrt%7B%5Cdfrac%7B0.04342975%7D%7B1000%7D%7D%7D)
![z = \dfrac{0.0045}{\sqrt{4.342975 \times 10^{-5}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cdfrac%7B0.0045%7D%7B%5Csqrt%7B4.342975%20%5Ctimes%2010%5E%7B-5%7D%7D%7D)
z = 0.6828
The p-value = P(z ≤ 0.6828)
From the z tables
p-value = 0.7526
Thus, the probability that at most 5% of the items in a given batch will be defective = 0.7526
The z - score for this distribution for at least 85% of the items is;
![z = \dfrac{0.85 - 0.9545}{\sqrt{\dfrac{0.0455\times 0.9545}{1000}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cdfrac%7B0.85%20-%200.9545%7D%7B%5Csqrt%7B%5Cdfrac%7B0.0455%5Ctimes%200.9545%7D%7B1000%7D%7D%7D)
![z = \dfrac{-0.1045}{\sqrt{\dfrac{0.04342975}{1000}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cdfrac%7B-0.1045%7D%7B%5Csqrt%7B%5Cdfrac%7B0.04342975%7D%7B1000%7D%7D%7D)
z = −15.86
p-value = P(z ≥ -15.86)
p-value = 1 - P(z < -15.86)
p-value = 1 - 0
p-value = 1
Thus, the probability that at least 85% of these items in a given batch will be acceptable = 1