# economyst check!

posted by
**stats!** on
.

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.