Answer:
The probability that more than 10 parts will be defective is 0.99989.
Step-by-step explanation:
Let <em>X</em> = a part in the shipment is defective.
The probability of a defective part is, P (Defect) = <em>p</em> = 0.03.
The size of the sample is: <em>n</em> = 1000.
Thus, the random variable
.
But the sample size is very large.
The binomial distribution can be approximated by the Normal distribution if the following conditions are satisfied:
- np ≥ 10
- n (1 - p) ≥ 10
Check the conditions:
![np=1000\times0.03=30>10\\n(1-p)=1000\times(1-0.03)=970>10](https://tex.z-dn.net/?f=np%3D1000%5Ctimes0.03%3D30%3E10%5C%5Cn%281-p%29%3D1000%5Ctimes%281-0.03%29%3D970%3E10)
Thus, the binomial distribution can be approximated by the Normal distribution.
The sample proportion (<em>p</em>) follows a normal distribution.
Mean: ![\mu_{p}=0.03](https://tex.z-dn.net/?f=%5Cmu_%7Bp%7D%3D0.03)
Standard deviation: ![\sigma_{p}=\sqrt{\frac{p(1-p)}{n} } =\sqrt{\frac{0.03(1-0.03)}{1000} } =0.0054](https://tex.z-dn.net/?f=%5Csigma_%7Bp%7D%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%20%7D%20%3D%5Csqrt%7B%5Cfrac%7B0.03%281-0.03%29%7D%7B1000%7D%20%7D%20%3D0.0054)
Compute the probability that there will be more than 10 defective parts in this shipment as follows:
The proportion of 10 defectives in 1000 parts is: ![p=\frac{10}{1000}=0.01](https://tex.z-dn.net/?f=p%3D%5Cfrac%7B10%7D%7B1000%7D%3D0.01)
The probability is:
![P(p>0.01)=P(\frac{p-\mu_{p}}{\sigma_{p}}> \frac{0.01-0.03}{0.0054}) =P(Z>-3.704)=P(Z](https://tex.z-dn.net/?f=P%28p%3E0.01%29%3DP%28%5Cfrac%7Bp-%5Cmu_%7Bp%7D%7D%7B%5Csigma_%7Bp%7D%7D%3E%20%5Cfrac%7B0.01-0.03%7D%7B0.0054%7D%29%20%3DP%28Z%3E-3.704%29%3DP%28Z%3C3.704%29)
Use the standard normal table for the probability.
![P(p>0.01)=P(Z](https://tex.z-dn.net/?f=P%28p%3E0.01%29%3DP%28Z%3C3.704%29%3D0.99989)
Thus, the probability that more than 10 parts will be defective is 0.99989.