# Using normal distribution to approximate binomial distribution (continuity correction)

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?

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