Among U.S. households, 24% have telephone answering machines. If a telemarketing campaign involves 2,500 households, find the probability that more than 650 have answering machines. The distribution of households with an answering machine is really a binomial distribution: either you have one or you don't, there is a flat chance of having one (0.24) and all households are independent.
What you are going to do is to approximate the result with a normal distribution:
If a variable has a binomial distribution B(n,p), and you have both np and n(1-p) greater than 10, you can estimate your probability with a normal distribution N(np, (鈭歯p(1-p))虏)
In your case, p= 0.24, n = 2500
np = 600, n(1-p) = 1900 You should be safe.
The normal distribution you are going to use has a mean of np=600 and a standard deviation of 鈭?np(1-p)) = 21.35
You are trying to find P(N>650) which is 1 - P(N鈮?50)
You have to apply a continuity correction to take into account the fact you are going from a discrete distribution (binomial) to a continuous distribution (normal). You add 0.5 to your upper bound as a correction:
P(N>650) = 1- P(N鈮?50.5)
P(N鈮?50.5) = 桅((650.5-600)/21.35) = 桅(2.365) = 0.991
Finally, P(N>650) = 1 -0.991 = 0.009, or 0.9%
You can also program the binomial distribution on your calculator and do it the hard way:
P(N>650) = 危(k= 651 to 2500)P(N=k)
where P(N=k) = 0.24^k*0.76^(2500- k)*2500!/(k!(2500 -k)!)
I get P(N>650) = 0.00946
The approximation was quite good. Let Xb be the number of answering machines. Xb has the binomial distribution with n = 2500 trials and success probability p = 0.24
In general, if X has the binomial distribution with n trials and a success probability of p then
P[Xb = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[Xb = x] = 0 for any other value of x.
To use the normal approximation to the binomial you must first validate that you have more than 10 expected successes and 10 expected failures. In other words, you need to have n * p > 10 and n * (1-p) > 10.
Some authors will say you only need 5 expected successes and 5 expected failures to use this approximation. If you are working towards the center of the distribution then this condition should be sufficient. However, the approximations in the tails of the distribution will be weaker especially if the success probability is low or high. Using 10 expected successes and 10 expected failures is a more conservative approach but will allow for better approximations especially when p is small or p is large.
In this case you have:
n * p = 2500 * 0.24 = 600 expected success
n * (1 - p) = 2500 * 0.76 = 1900 expected failures
We have checked and confirmed that there are enough expected successes and expected failures. Now we can move on to the rest of the work.
If Xb ~ Binomial(n, p) then we can approximate probabilities using the normal distribution where Xn is normal with mean 渭 = n * p, variance 蟽虏 = n * p * (1-p), and standard deviation 蟽
Xb ~ Binomial(n = 2500 , p = 0.24 )
Xn ~ Normal( 渭 = 600 , 蟽虏 = 456 )
Xn ~ Normal( 渭 = 600 , 蟽 = 21.35416 )
I have noted two different notations for the Normal distribution, one using the variance and one using the standard deviation. In most textbooks and in most of the literature, the parameters used to denote the Normal distribution are the mean and the variance. In most software programs, the standard notation is to use the mean and the standard deviation.
The probabilities are approximated using a continuity correction. We need to use a continuity correction because we are estimating discrete probabilities with a continuous distribution. The best way to make sure you use the correct continuity correction is to draw out a small histogram of the binomial distribution and shade in the values you need. The continuity correction accounts for the area of the boxes that would be missing or would be extra under the normal curve.
P( Xb < x) 鈮?P( Xn < (x - 0.5) )
P( Xb > x) 鈮?P( Xn > (x + 0.5) )
P( Xb 鈮?x) 鈮?P( Xn 鈮?(x + 0.5) )
P( Xb 鈮?x) 鈮?P( Xn 鈮?(x - 0.5) )
P( Xb = x) 鈮?P( (x - 0.5) < Xn < (x + 0.5) )
P( a 鈮?Xb 鈮?b ) 鈮?P( (a - 0.5) < Xn < (b + 0.5) )
P( a 鈮?Xb < b ) 鈮?P( (a - 0.5) < Xn < (b - 0.5) )
P( a < Xb 鈮?b ) 鈮?P( (a + 0.5) < Xn < (b + 0.5) )
P( a < Xb < b ) 鈮?P( (a + 0.5) < Xn < (b - 0.5) )
In the work that follows Xb has the binomial distribution, Xn has the normal distribution and Z has the standard normal distribution.
Remember that for any normal random variable Xn, you can transform it into standard units via: Z = (Xn - 渭 ) / 蟽
P( Xb > 650 ) =
2500
鈭?P(Xb = x) = 0.009460844
x = 651
鈮?P( Xn > 650.5 )
= P( Z > ( 650.5 - 600 ) / 21.35416 )
= P( Z > 2.364879 )
= 0.009017977 |
