For a factory having 1000 devices, after inspection 20 devices were defectives, find the probability distribution and the cumulative probability distributions if 5 devices were selected ?
+118629 For a factory having 1000 devices, after inspection 20 devices were defectives, find the probability distribution and the cumulative probability distributions if 5 devices were selected ?
P(defective)= 20/1000 = 0.02
This is a strangly worded question - maybe that is because I don't fully understand it.
Anyway there is a binomial calculator to get what you want maybe.
http://stattrek.com/online-calculator/binomial.aspx
ans
This is the formula.
\(P(\text{n defectives in 5 trials})=\binom{5}{n}*0.02^n*0.98^{(5-n)}\)
\(P(\text{n or less then n defectives in 5 trials})=\displaystyle\sum_{k=0}^{k=n}\: \left[\binom{5}{k}*0.02^k*0.98^{(5-k)}\right] \)
