David, the normal approximation applies a mean μ and standard deviation σ to a binomial distribution (Which is based on sample size n and probability p). Basically the approximated mean would be np and the standard deviation would be sqrt(npq) where q is 1-p. Last but not least we are given the sample point x=47. So now we can calculate a z value:
(x-np)/√npq=(47-42.46)/5.75=.7896
Which when referred to a z table, equates to: .7852 which means everything from 0 to 47 on the distribution with mean 42.46 and standard deviation of 5.75 would be 78.52% of the distribution.
Hope that helps!