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Hi Carolyn,

Note the question says to use the normal approximation to calculate the probability that P(x > or = 17) for n=30, p=0.50. The normal approximation is the key phrase.

The approach is to use the normal approximation instead of the binomial distribution. This is a review topic, but since it is optional, not everyone sees this in Stats I. Here is what you do:

P( x >=17) for n = 30, p = 0.50 This means we want to first find the mean for the binomial random variable, mean = n*P = 0.5*30 = 15 (this you should have seen in Stats I). The standard deviation is sigma = root(n*p*q) = root(30*0.5*0.5) = 2.7386. Now, we use the normal curve to find the probability that x >= 17 (this part is also review). Usually, we would get the z-score for x = 17 by plugging it into z = (x - mean)/sigma. However, we can't just use the 17. We are approximating the binomial curve by the bell curve. A discrete distribution like the binomial curve actually uses rectangles and those rectangles of probability actually range from a half point before the value and a half point above (16.5 until 17.5). This means we can really improve the approximation by using 16.5 or above. This way as we sweep up the probability from 17 and above we don't miss that little bit of the rectangle for 17 that starts at 16.5. This is easier to explain when we have a board to show this on, but hopefully it is clear. This adding or subtracting 0.5 from the x value is called continuity correction. You can just memorize the several cases if need be, but I prefer you think it through. If you did memorize it, just remember if you are doing greater than or equal to, you need to subtract 0.5. If you are doing less than or equal to, you need to add 0.5.

Now, this means that we will find z as, z = (x - mean)/sigma = (16.5 - 15)/2.7386 = 0.55. Now, use the z-table to look up 0.55, you will get 0.2088. This is from the center until 0.55, so we need to subtract 0.2088 from 0.5000 which gives us the solution = 0.2912.

If you were good with the z-table in your stats I class, this part should be easy for you.

Professor McGuckian