Find all the real square roots of 0.0009

Let A = 0.0009, it is equivalent to A = 9^-4 =  (9^-2)^2
so sqrt A =sqrt (9^-2)^2 = (9^-2)  
he real square roots of 0.0009 is 0.09

In this situation, 
n=50, p=1/20, q=(1-p)=19/20, and npq=19/8=2.4

We would like np and npq to be a large number, at least greater than 10.
The normal approximation can always be applied, but the result will be very approximate, depending on the values of np and npq.

Situations are favourable for the normal approximation when p is around 0.5, say between 0.3 and 0.7, and n>30.

"Normal approximation" is using normal probability distribution to approximate the binomial distribution, when n is large (greater than 70) or exceeds the capacity of most hand-held calculators.  The binomial distribution can be used if the following conditions are met:
1. Bernoulli trials, i.e. exactly two possible outcomes.2. Number of trials is known before and constant throughout the experiment, i.e. independent of outcomes.3. All trials are independent of each other.4. Probability of success is known, and remain constant throughout trials.
If all criteria are satisfied, we can model with binomial distribution, where the probability of x successes out of N trials each with probability of success p is given byP(x)=C(N,x)(p^x)(1-p)^(N-x)and,C(N,x) is number of combinations of selecting x objects out of N.
The mean is np, and variance is npq.

For the given situation, np=2.5, npq=2.375, so standard deviation=sqrt(2.375)=1.54.