So what we're interested in is the sum of probabilities

of each one of these outcomes 70 through 245.

We can calculate each one of these probabilities using the binomial formula

and add them up, but that really does not sound like fun.

This is where the resemblance between the binomial

distribution and the normal distribution comes in very handy.

The blue-shaded area of interest can just as

well be calculated as the area under the smooth

normal curve that closely resembles the more jagged binomial distribution.

Because calculating a shaded area under the normal

curve is a much simpler task than calculating individual

binomial probabilities for all of these outcomes and

adding them up, we might want to use that method.

To calculate a normal probability, we need a little

more information on the parameters of the normal distribution.

These can be estimated by the mean and the standard deviation of the original

binomial distribution. The mean is n times p, so that's 245

times 0.25, 61.25, and the standard deviation

is the square root of 245 times 0.25 times 0.75

Which comes out to be 6.78. So among 245 friends,

we expect 61.25 power users, give or take 6.78.

Given an observation, the mean, and the standard deviation, we

can calculate the area under the curve via a z score.

So the z score is going to be the observation 70 minus 61.25,

the mean, divided by 6.78, the standard deviation, which comes

out to be 1.29.

We can then find the probability of a z score being greater than 1.29, since

we shaded the area underneath the curve beyond the observation of interest.

So we want to take a look on our table to 1.29 as a z score, and in the

intersection of the row and the column of interest, we can see 0.9015.

The probability of obtaining

a z score greater than 1.29 is going to be one minus that probability from the table.

Why are we doing this one minus bit?

Well, because the table always gives us the percentile or the area under the

curve below the observed value and we want to find the complement of that.

Which comes out to be 0.0985. So there is a 9.85%

chance that an average Facebook user, with 245 friends,

has at least 70 friends who are considered power users.