According to a market research firm, 54% of all residential telephone numbers in Los Angeles are unlisted. A telephone sales firm uses random digit dialing equipment that dials residential numbers at random, whether or not they are listed in the telephone directory. The firm calls 512 numbers in Los Angeles.
(a) What are the mean and standard deviation of the proportion of unlisted numbers in the sample?
(b) What is the probability that at least half the numbers dialed are unlisted? (Remember to check that you can use the normal approximation.) Let Xb be the unlisted phone numbers. Xb has the binomial distribution with n = 512 trials and success probability p = 0.54
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 = 512 * 0.54 = 276.48 expected success
n * (1 - p) = 512 * 0.46 = 235.52 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 = 512 , p = 0.54 )
Xn ~ Normal( 渭 = 276.48 , 蟽虏 = 127.1808 )
Xn ~ Normal( 渭 = 276.48 , 蟽 = 11.27745 )
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 鈮?256 ) =
512
鈭?P(Xb = x) = 0.9684294
x = 256
鈮?P( Xn 鈮?255.5 )
= P( Z 鈮?( 255.5 - 276.48 ) / 11.27745 )
= P( Z 鈮?-1.860350 )
= 0.968582 |
