Question 937681: Studies have shown that 24% of all students in high school smoke. Suppose we happen to observe a random sample of 70 students in the cafeteria. What is the probability that they proportion of those 70 students is no more than 20%?
I know I need to solve for P, but I am lost as to how to get there with out the mu or sigma.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! In a population of size N, suppose that the probability of the occurrence of an event (dubbed a "success") is P; and the probability of the event's non-occurrence (dubbed a "failure") is Q. From this population, suppose that we draw a sample of size n. And finally, within this sample, we determine the proportion of successes p and failures q. In this way, we create a sampling distribution of the proportion.
We find that the mean of the sampling distribution of the proportion (mu of p) is equal to the probability of success in the population (P). And the standard error of the sampling distribution (sigma of p) is determined by the standard deviation of the population (sigma), the population size, and the sample size. These relationships are shown in the equations below:
mu of p = P and sigma of p = sigma * sqrt( 1/n - 1/N ) = sqrt[ PQ/n - PQ/N ]
where sigma = sqrt[ PQ ].
Note: When the population size is very large, the factor PQ/N is approximately equal to zero; and the standard deviation formula reduces to: sigma of p = sqrt( PQ/n ). You often see this formula in intro statistics texts.
now we can solve the problem
P = .24
Q = .76
n = 70
mu of p = .24
assume N is very large compared to n=70
sigma of p = square root ( (.24*.76) / 70 ) = 0.051046198
now we can calculate the z-value
z-value = (.20 - .24) / 0.051046198 = -0.78360391
consult z-value tables
P(X<.20) = 0.2177
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