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Normal approximation for Binomial Distribution. I am using the correct formula, but my answer is wrong. I think it may be due to round off. Attached image.
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\[P(X>= 208) = \frac{ 208 - 1000*0.193 }{ \sqrt{1000*0.193*0.807} } \rightarrow z=1.2019 \]
z of 1.2019 = 0.8853 (approx)\[P(x \ge208) = 1-0.8853 = 11.47\]
There was a similar problem and my answer was really close.
It looks correct
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The answer is 0.1071. I hate these lol. Thanks for confirming.
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