Question

Suppose that the probability of a defective part is 0.03. Suppose that you have a shipment of 1000 parts. What is the probability that more than 10 parts will be defective? Answer to at least five decimal places. You may find it easier to use excel than to use a calculator for this one.

Answer 1

Answer:

The probability that more than 10 parts will be defective is 0.99989.

Step-by-step explanation:

Let X = a part in the shipment is defective.

The probability of a defective part is, P (Defect) = p = 0.03.

The size of the sample is: n = 1000.

Thus, the random variable $X\sim Bin(1000, 0.03)$.

But the sample size is very large.

The binomial distribution can be approximated by the Normal distribution if the following conditions are satisfied:

1. np ≥ 10
2. n (1 - p) ≥ 10

Check the conditions:

$np=1000\times0.03=30>10\\n(1-p)=1000\times(1-0.03)=970>10$

Thus, the binomial distribution can be approximated by the Normal distribution.

The sample proportion (p) follows a normal distribution.

Mean: $\mu_{p}=0.03$

Standard deviation: $\sigma_{p}=\sqrt{\frac{p(1-p)}{n} } =\sqrt{\frac{0.03(1-0.03)}{1000} } =0.0054$

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: $p=\frac{10}{1000}=0.01$

The probability is:

$P(p>0.01)=P(\frac{p-\mu_{p}}{\sigma_{p}}> \frac{0.01-0.03}{0.0054}) =P(Z>-3.704)=P(Z<3.704)$

Use the standard normal table for the probability.

$P(p>0.01)=P(Z<3.704)=0.99989$

Thus, the probability that more than 10 parts will be defective is 0.99989.