I’m in high school and am learning sample proportions. I have encountered a doubt that i cannot answer myself.

Why do we use normal **approximation** for sample proportions of cases involving a binomial distribution?

Why approximate? We have a binomial distribution, isn’t it more accurate to just use this?

For example,

A company employs a sales team of 20 people, consisting of 12 men and 8 women. 5 sales people are to be selected at random to attend an important conference. Determine the probability that the proportion of men, in a random sample of 5 selected from the sales team, is greater than 0.7.

To solve this question, my teachers would say use normal distribution with X being N ~ (3, 0.048) to calculate the probabilty of (X>0.7) in my calculator.

My problem here is: wouldn’t it make more sense to calculate using binomial distribution with X being Bin ~ (5,0.6) then, as 0.7*5 is 3.5 round down to 3 as you can’t have 3.5 men, calculating probability of (X>3) in my calculator.

Can someone please clarify this? ðŸ˜€

Thanks