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 by[tex]P(x)=C(n,x)p^x(1-p)^{n-x}[/tex]where[tex]C(n,x)=\frac{n!}{x!(n-x)!}[/tex]
[tex]P(\ge20)[/tex]
[tex]=\sum_{i=20}^{250}P(i)[/tex]
[tex]=\sum_{i=20}^{250}C(250,i)p^i(1-p)^(250-i)[/tex]
=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- μ )/ σ )
=1-P(Z< (19.5-17.5)/4.0342)
=1-P(Z<0.49576)
=1-0.689967
=0.3100 (using normal approximation)
