A manufacturer of toys recalled one of their products when it was discovered that 30% of the toys had a defective part. In a shipment of 4000 of these toys, what is the probability that fewer than 1150 are defective?

Question

kropot72 · Answer

The binomial distribution applies to this problem. The binomial distribution can be approximated by the normal distribution where n is large and p is neither small nor near 1.
The mean = np = 4000 * 0.3
The standard deviation is$$\sqrt{npq}=\sqrt{4000\times .3\times .7}$$
The next step is to find the z-score where X = 1150. Can you do this?

anonymous · Answer

.... it's not making much sense to me...

anonymous · Answer

.... it's not making much sense to me...

anonymous · Answer

.... it's not making much sense to me...

anonymous · Answer

.... it's not making much sense to me...

kropot72 · Answer

What part of my explanation doesn't make sense?