Answer:
The number of defective bulbs in the shipment should be around 15, give or take 4.
Explanation:
For each bulb, there are only two possible outcomes. Either it is defective, or it is not. The probability of a bulb being defective is independent of any other bulb. This means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
Probability of exactly x successes on n repeated trials, with p probability.
The expected value of the binomial distribution is:
![E(X) = np](https://img.qammunity.org/2022/formulas/mathematics/college/vhithkjh7varsjyjym1v6ct4sm4mej9im1.png)
The standard deviation of the binomial distribution is:
![√(V(X)) = √(np(1-p))](https://img.qammunity.org/2022/formulas/mathematics/college/e69rpeoj1vt09gh26fkrtaiqmha25fl1ev.png)
Suppose the probability at a light bulb factory of a bulb being defective is 0.11
This means that
![p = 0.11](https://img.qammunity.org/2022/formulas/mathematics/college/8gr7btlqvtjfunp1xthmth6y4szga7smhj.png)
Shipment of 133 bulbs:
This means that
![n = 133](https://img.qammunity.org/2022/formulas/mathematics/college/4x6jlptzp0vorr9ctaj7voz7gnhof8emtc.png)
Mean and standard deviation:
![E(X) = np = 133*0.11 = 14.63](https://img.qammunity.org/2022/formulas/mathematics/college/7vjbj02lq5wyidg9zazshsfy8v1ix1vghk.png)
![√(V(X)) = √(np(1-p)) = √(133*0.11*0.89) = 3.61](https://img.qammunity.org/2022/formulas/mathematics/college/b0wk5e03ehepk3y4l2tfpvry1ravst87fx.png)
Rounding to the nearest integers:
The number of defective bulbs in the shipment should be around 15, give or take 4.