Saturday

December 27, 2014

December 27, 2014

Posted by **stats!** on Thursday, June 7, 2007 at 4:53pm.

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 this Question**

**Related Questions**

Reposted: Math (for "google" - In many families both parents work, as a result ...

English - I really need help with my writing. I need some to proof read a paper...

Socials - Imagine that you are a factory owner in eighteenth-century britain. ...

Working with Parents - Why must parents be involved in their child's day care ...

computers in day care - Which of the following statements about children and ...

Health of young children - Why is the incubation stage of an illness of concern ...

English - Volunteering at a day-care center. I have worked at a day-care center...

GEOMETRY/PROP 2 - 72% of employees bring their children to day-care centers ...

geometry - Need help with a word problem - Carrie collected can food for a ...

early childhood education - voters have been historically nwilling to underwrite...