Question 698345: Random Sampling:According to the 2000 U.S.Census the city of Miami,Florida has a population of 362,470 of whom 238,351 are Latino or hispanic. If 500 citizens of Miami are selected at random, what is the probability that between 300 and 340(inclusive)of them will be latino or hispanic?
Answer by Positive_EV(69)   (Show Source):
You can put this solution on YOUR website! The population proportion of Latino or Hispanic residents in Miami is 238351/362470 = .65757. In order to solve this problem, since there is a large sample, you can approximate what would be a pretty nasty binomial calculation with the Central Limit Theorem. That is, for large samples, the distribution will tend to be approximately normal.
The mean of the number of people out of 500 who are Latino is (.65757)*(500) = 328.785. The standard deviation of one person being selected at random and being a Latino (1 if Latino, 0 if not) is  =  .
Knowing the mean and standard deviation, we can calculate the appropriate Z-scores with the normal distribution. However, since the binomial distribution takes on discrete values and the normal distribution is continuous, the continuity correction must be applied to the range of values before calculating Z to take this into account. This involves expanding the range of the desired probability by .5 in both directions, so we are now interested in the probability that between 299.5 and 340.5 people surveyed are Latino.
The Z-score is calculated as  , where X will be the two bounds we are looking to find the Z-scores for, mu = the population mean = 328.785, sigma = the standard deviation = .47452, and n = the number of people surveyed = 500. Calculating the Z-values for 299.5 and 340.5:
So you want the probability that (-2.76 < Z < 1.10). Looking at a Z-table such as the one found at http://lilt.ilstu.edu/dasacke/eco148/ztable.htm, we see that Z(-2.76) = .0029, and Z(1.10) = .8643. A value on the Z table gives the probability that, given the population mean, a value is less than the value found. Thus, the probability that the number of people surveyed is less than 340 but greater than 300 is .8643 - .0029 = .8614.
Note that some teachers don't use the continuity correction. In this case, repeat the calculations for the Z values above, only use X = 300 and 340 instead of 299.5 and 340.5.
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