Word Problem?

Question

Suppose that a population of adults is surveyed to number the measure the number of hours per week spent on home computers. In the survery, the number of hours is normally distibuted with a mean of 7 hours and a standard deviation of 1 hour.

Find the % of adults that spend more than 4.9 hours on a home computer per week.

If 150 adults are randomley selected, about how many would you expect to say they spen less than 5.4 hours per week on a home computer?

Answer 1

This is a simple statistics problem based on the properties of the Gaussian distribution.

The fraction of the population that is within n standard deviations of the mean given a Gaussian distribution is given by erf(n/sqrt(2)), where "erf" stands for the "error function":

http://en.wikipedia.org/wiki/Error_function
http://en.wikipedia.org/wiki/Normal_distribution

Some values for erf(n/sqrt(2)) are:

1 -> 0.682689492137
2 -> 0.954499736104
3 -> 0.997300203937
4 -> 0.999936657516
5 -> 0.999999426697
6 -> 0.999999998027

That means that 68.3% are within one standard deviation of the mean, with the remaining 31.7% split into half more than one standard deviation below the mean and half more than one standard deviation above the mean.

Similarly, 95% are within two standard deviations of the mean and 99.9999998% are within 6 standard deviations of the mean.

Here is a tabulation of more values:
coe.montana.edu/composites/.../Table of Error Function Values.pdf

(Some statistical calculators have this function; there are other tables, etc.)

So, now let's look at the problems.

4.9 hours is 2.1 hours away from the mean of 7 hours, or 2.1 standard deviations below. So the fraction that spends more than 4.9 hours is the fraction that is within 2.1 standard deviations, plus the half of the remainder, the people that spend more than 9.1 hours/week.

So:
let E = erf(2.1/sqrt(2))
let C = 1 - E
then the fraction we want is E + C/2

For the second question we first find the number of standard deviations represented by 5.4 hours, compute the corresponding E value, and then the corresponding C value.

C/2 is the fraction that spend less than the n standard deviations at their computer so (150)(C/2) is the expected number of people.

BTW, note that this is only a theoretical problem. It simply isn't possible for the number of hours to be normally distributed. A true normal distribution is symmetric and extends to infinity on both sides of the mean.

In this case, people can spend more than 14 hours/week at their computer, but no one can spend less than 0 hours/week.

But people use the Gaussian distribution anyway, just because it is so easy to use. Beware.