Question

Stats help please!
Suppose 4-year-olds in a certain country average 3 hours a day unsupervised and that most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.8 hours and the amount of time spent alone is normally distributed. We randomly survey one 4-year-old living in a rural area. We are interested in the amount of time the child spends alone per day.
80% of the children spend at least how long per day unsupervised?

Answer 1

i do not know this but i would just like you to know someone. @dumbcow is a statistics specialist of OpenStudy. I suggest you fan him so you can see when he's online so you can ask for statistics help from him :)

Answer 2

right now,he's not online sadly

Answer 3

this uses a standard normal distribution curve, to find Z-scores you need a standard normal table to look up probabilities
http://www.sjsu.edu/faculty/gerstman/EpiInfo/z-table.htm
the goal is to find the amount of hours where 80% of children spend more than that amount unsupervised

the formula for finding Z-score is
\[Z = \frac{X-mean}{std dev} = \frac{X-3}{1.8}\]
however, we don't know X, that is the answer or the desired amount of time
to determine what Z-score to use, look at table and look up probability of 80% or 0.8
From the table the closest value to 0.8 is 0.7995 , this corresponds to Z-score of 0.84
but since we know Z must be neg from the bell curve drawing...use Z = -0.84
Now solve for X
\[-0.84 = \frac{X-3}{1.8}\]
\[X = 1.488\]
done! 80% of children spend at least 1.488 hours unsupervised