Could anyone help with the following problem?
My guts is telling me that the answer to part (a) is a normal distribution. Mainly, because I can't see where a uniform distribution would fit in this problem.
However, I'm confused as to how to determine the probability in part (b). I know that for a given mean and standard deviation, we can find a z score and then use a table look up to find the value. But in this problem I'm not given mean or standard deviation. Is there a way I should be determining these values through the given information?
Ad a) The random variable is binomial distributed, $X \sim B(400,0.4)$.
Ad b) Now you can approximate the binomial distribution by the normal distribution.
You take the mean and the standard deviation of the binomial distributed random variable. $\mu=0.4\cdot 170$ and $ \sigma=\sqrt{ \sigma ^2 } =\sqrt{n\cdot p \cdot (1-p)}=\sqrt{400\cdot 0.4\cdot 0.6}$
$P(X \leq x) \approx \Phi \left( \frac{x+0.5-n \cdot p}{\sqrt{n \cdot p \cdot (1-p)}} \right)$
$P(X < 170)=P(X \leq 169) \approx \Phi \left( \frac{169+0.5-400 \cdot 0.4}{\sqrt{400\cdot 0.4\cdot 0.6}} \right)$
$X \sim B(400,0.4)$
I allready have added the continuity correction, which is 0.5 in the numerator. But I think it is not really neccessary, because n is quiet large.