# AP Stat question- bcdf or z test??

<p>in barrons there is the following question (page 293): suppose that 70% of all dialysis patients will survive for at least 5 years. In a SRS of 100 new patients, what is the probability that the proportion surviving for at least 5 years will exceed 80%?</p>

<p>barrons calculates the single proportion standard deviation and then evaluates using ncdf. Can this be done using bcdf? I think it would make sense to do 1-bcdf(100, .7, 80), but I don't get the same answer. Can someone please explain this to me?</p>

<p>P = .7
N = 100</p>

<p>std = sqrt ( (P)(1-P) / N) = .0458</p>

<p>ncdf ( .8, 999, .7, .0458) = .0145 = 1.45%</p>

<p>yeah but I'm asking why can't it be done using bcdf. How would I even know which method to use?</p>

<p>You use bcdf for things like coin flips. The question will usually ask like, out of 25 coin flips, what is probability you get more than 15 heads?</p>

<p>1 - bcdf( 25, .5, 15) = 11.5%</p>

<p>two outcomes: die or survive
fixed probability: .7
assuming normality and independence, we should be able to apply bcdf...</p>

<p>the question seems akin to saying out of 100 coin flips what is the probability of more than 80 heads.</p>

<p>You're taking an SRS, which is approximately normal. Thus, you can't use binomcdf, which is based on counts. However, if you calculate the standard deviation - sqrt(pqn) - and mean - np - of the binomial distribution, you'll find that if you use those with normalcdf, you will get 0.0145 as your answer still. The situation is binomial, but since you're taking an SRS, you can't use the binomial distribution function.</p>

<p>****, if we can't use binomial for this I don't know what to do.</p>

<p>mtv22 - there are two ways to approach this. One, as a sampling proportion:</p>

<p>Standard deviation = sqrt(pq/n) = sqrt(.7*.3/100) = 0.045826
Mean = p = 0.70
normalcdf(0.8,1,0.70,0.045826) = 0.01455</p>

<p>Second, as a normal binomial approximation (no clue if that's the right terminology or not):</p>

<p>Standard deviation = sqrt(pqn) = sqrt(0.7 * 0.3 * 100) = 4.5826
Mean = np = 100 * 0.70 = 70
normalcdf(80,100,70,4.5826) = 0.01455</p>

<p>so if there is an SRS involved, the solution cannot use bcdf?</p>