If the probability of hearing a "yes" to a campaign contribution drive is 0.07, and 250 requests are made, what is the probability of getting at least 20 positive responses?

Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n=250)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p=0.07)4. All trials are random and independent of the others
The number of successes, x, is then given bywhere



=0.30069   to 5 decimal places using binomial distribution

Note: it is possible to actually do the summation (as is done here), or refer to binomial tables/software, or to use the normal approximation, with continuity correction, as shown below.

mean number of responses = np = 250*0.07 = 17.5
variance of responses  = npq = np(1-p) = 250(0.07)(0.93)=16.275
standard deviation = sqrt(16.275)=4.0342
P(X>=19.5)
=1-P(X<19.5)
=1-P(Z<(X- &mu; )/ &sigma; )
=1-P(Z< (19.5-17.5)/4.0342)
=1-P(Z<0.49576)
=1-0.689967
=0.3100   (using normal approximation)