In many families both parents work, as a result there is increasing need for day care. Data was collected; and one year in Canada approximately 32% of children aged 0 to 11 years were in day care for at least 20h per week.

Question

In many families both parents work, as a result there is increasing need for day care. Data was collected; and one year in Canada approximately 32% of children aged 0 to 11 years were in day care for at least 20h per week. 
What is the probability, in a random poll of 60 children from the ages of 0 to 11 that more than 15 children are in day care at least 20h per week? 
mean = .32*60 
mean=19.2 
standard deviation: 
of sqrt(n*p*q) 
= sqrt(60*.32*.68) 
= 3.613. 
therfore: 
19.2 -15.5 
= 3.7

my question is were did the 6.13 come from?

What is the probability in a random poll of 60 children that fewer than 20 are in a day care at least 20h per week? 
mean = .32*60 
mean=19.2 
standard deviation: 
of sqrt(n*p*q) 
= sqrt(60*.32*.68) 
= 3.613. 
19.5 -19.2 
= 0.3 
oksy from here what next? 
divide it by 6.13? thanks! 
my question is were did the 6.13 come from? 
thanks!

6.13??? 
The estimated standard deviation is 3.613 -- do you mean this??

Then for part 2, divide .3 by 3.613 = 0.08. Then look up 0.08 in your cumulative normal distribution table. I get .5319. Ergo, I would say there is a 53% chance you would see fewer than 20.