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.

Answer 1

I apologize for the confusion. The value of 6.13 was a typo, and it should have been 3.613 as you correctly pointed out.

To calculate the probability that more than 15 children are in day care for at least 20 hours per week in a random poll of 60 children, you first need to calculate the z-score.

Z-score = (x - mean) / standard deviation

where x is the number of children (15 in this case), mean is the mean value (19.2), and standard deviation is the estimated standard deviation (3.613).

Z-score = (15 - 19.2) / 3.613 = -1.16

Once you have the z-score, you can use the z-score table or a statistical calculator to find the probability associated with that z-score. In this case, since we want to calculate the probability of more than 15 children, you need to find the cumulative probability to the right of the z-score.

For the probability that fewer than 20 children are in day care for at least 20 hours per week, you would follow a similar process.

Z-score = (20 - 19.2) / 3.613 = 0.22

Again, you would find the cumulative probability to the left of the z-score. In this case, the probability is 0.4224 or approximately 42.24%.

I hope this clears up the confusion. Please let me know if you have any further questions!