economyst check!

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
standard deviation:
of sqrt(n*p*q)
= sqrt(60*.32*.68)
= 3.613.
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
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?

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.

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