Answer:
The probability that at least 14 of the chips will be defective is 0.6664.
Explanation:
The complete question is:
A manufacturing process produces semiconductor chips with a known failure rate of 5.4%. If a random sample of 300 chips is selected, approximate the probability that at least 14 will be defective. Use the normal approximation to the binomial with a correction for continuity .
Solution:
Let X = number of defective chips.
The probability that a chip is defective is, p = 0.054.
A random sample of n = 300 chips is selected.
A chip is defective or not is independent of the other chips.
The random variable X follows a Binomial distribution with parameters n = 300 and p = 0.054.
But the sample selected is too large.
So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:
- np ≥ 10
- n(1 - p) ≥ 10
Check the conditions as follows:
![np=300* 0.054=16.2>10\\n(1-p)=300* (1-0.054)=283.8>10](https://img.qammunity.org/2021/formulas/mathematics/high-school/4x8q8uip0dfhiqxjdtnt1k2c6qt1tv8bu9.png)
Thus, a Normal approximation to binomial can be applied.
So,
.
Compute the probability that at least 14 of the 300 chips will be defective as follows:
Use continuity correction:
P (X ≥ 14) = P (X > 14 + 0.50)
= P (X > 14.50)
![=P((X-\mu)/(\sigma)>(14.50-16.20)/(√(15.3252)))](https://img.qammunity.org/2021/formulas/mathematics/high-school/k2vfc3vghxpathq86deytl6rn5r6k7dp4e.png)
![=P(Z>-0.43)\\=P(Z<0.43)\\=0.6664](https://img.qammunity.org/2021/formulas/mathematics/high-school/bea50uof90i2o5crtth67g7y35g13zu1ov.png)
*Use a z-table for the probability.
Thus, the probability that at least 14 of the chips will be defective is 0.6664.