Question

Given that the probability of a trial is 0.6 and 25 trials are conducted, by approximating the distribution as normal what is the probability that between 14 and 17 trials are successful?

Answer 1

here for binomial distribution parameter n=25 and p=0.6

here mean of distribution=μ=np=		15.0000
and standard deviation σ=sqrt(np(1-p))=		2.4495
for normal distribution z score =(X-μ)/σx
since np and n(1-p) both are greater than 5, we can use normal approximation of binomial distribution
therefore from normal approximation of binomial distribution and continuity correction:

probability that between 14 and 17 trials are successful (inclusive) :

probability =P(13.5<X<17.5)=P((13.5-15)/2.449)<Z<(17.5-15)/2.449)=P(-0.61<Z<1.02)=0.8461-0.2709=0.5752

(Note: if 14 and 17 are exclusive: P(14.5 <X<16.5) =0.3084)