Is the Normal Distribution a Good Approximation for Binomial Probabilities?

[StatsCat33]

Homework Statement

According to a survey, only 20% of customers who visited the website on a major retail store made a purchase. Random samples of 75 are selected. 90% of the samples will have less than what percentage of customers who will make a purchase after visiting the website? (Round 2 decimal places)

Relevant Equations

SD = sqrt (p(1-p)/n)

1. I first started by finding the Standard Deviation and got it to be SD = sqrt((.15)(.85)/75) = .0412

2. I then thought it was was a probability they wanted me to find and deal with a normal distribution.
I found the normalcdf (-1000000000,.9,.15..0412) and I got 1. I assumed that was wrong and then stopped.

 

[StatsCat33]

Now that I think about it more, I think this is an inverse norm problem. I think I need to say the mean is .15 the SD = .0462 and then the invNorm of it being less than 90% would be 25.92%

 

[FactChecker]

The standard deviation of a general PDF is not a good way to determine probabilities. A "rule of thumb" is that the normal distribution is a good approximation of a binomial if both pn and qn=(1-p)n are greater than 5 (there is more to it than that which you might want to check). In your case, pn=.2*75=15 and (1-p)*75=60. Approximate it with $N$(np, np(1-p)). I can't figure out what you used.