Problem: Each item produced by a certain manufacturer is, independently, of acceptable quality with probability .95. Approximate the probability that at most 10 of the next 150 items produced are unacceptable.
The following is not a solution to the problem. I'm simply trying to show that normal distribution truly approximate binomial distribution. Binomial approach of having exactly 10 defective items
$\binom{150}{10} (0.05)^{10}(0.95)^{140} \approx 0.0869$
however, when I tried finding the cumulative probability(with continuity correction) using normal distribution:
between $\frac{(10.5-7.5)}{\sqrt{7.125}}$ and $\frac{(9.5-7.5)}{\sqrt{7.125}}$ (continuity correction), I get an answer $\approx$ $0.09632$. which is quite different from $0.0869$. Did i do something wrong? or is this the way it's supposed to be?
