How do I find the probability ?

Question

"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"

I need help finding the probability that the child spends less than 1 hour per day unsupervised. (I have to write the probability statement and find P)

Answer 1

Normal distribution ("bell curve") tells us what the probability distribution of "hours unsupervised" looks like. The probability relates to the position of the required value on the curve.

For example, the mean (Time = 3 hours unsupervised) is exactly in the middle, and P(T <= 3 hours) = 50%.

You might remember that one standard deviation includes a further 34% of the distribution, so we can say that P(1.2 hours <= T <= 4.8 hours) = 68% and P(T <= 1.2 hours) = 16% and P(T >= 4.8 hours) = 16%.

We need P(T <= 1 hour) which you can see must be slightly more than 1 SD, and hence will be slightly less than 16% (maybe 14% or 15% as a guesstimate).

To find the exact value, we need a Z score, and Z score table.

The Z score just converts our required hours into a number of standard deviations.

Z = (required value - mean) / SD

In our case, Z = (1 - 3) / 1.8 = -2/1.8 = -1.1111

A negative Z score means we are left of the mean, a positive score means we are to the right.

Looking this up in a Z Score table we find the value of 0.1335

Hence our required probability is 13.35% (in line with our guesstimate).

Note: Z for mean = (3-3) /1.8 = 0, and Z score for 0 is 0.5000 = 50% or exactly half of curve to left of middle.