What is the probability that the child spends less than 3.3 hours per day unsupervised.

Question

On the planet of Mercury, 4-year-olds average 3 hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.3 hours and the amount of time spent alone is normally distributed. We randomly survey one Mercurian 4-year-old living in a rural area. We are interested in the amount of time X the child spends alone per day. (Source: San Jose Mercury News) Round all answers to 4 decimal places where possible.

Answer 1

Maria,

So the question gives you a lot of crucial information:

1. the mean or μ of the time spent unsupervised of 3 hrs/day
2. the standard deviation or σ of 1.3 hrs/day
3. and finally, that the time spent unsupervised is normally distributed

If we are surveying 1 child then we are curious of answering a probability question about a variable X ~ N(3, 1.3) (read as X follows a normal distribution with mean 3 and standard deviation 1.3).

Your questions is P(X < 3.3). So on the bell curve the area to the left.

To make it easier let's transform it into Z-score instead. Z ~ N(0, 1) so to go from X -> Z we use:

Z = (X - μ) / σ

Plugging in:

Z = (3.3 - 3) / 1.3 ≈ 0.23

So our question becomes:

P(Z < 0.23)

We can use a table or code and we find that:

P(Z < 0.23) ≈ 0.5910