The dean of a statistics department in a certain university believes that 65% of the department’s graduate positions are given to international students. A random sample of 64 graduate assistants is taken. What is the probability that the sample proportion of graduate assistants who are international students will be between 0.6 and 0.75?
Solution
_{} = [p ( 1 - p ) / n] = [(0.65 * 0.35) / 64 ] = 0.0596
= P[(0.6 - 0.65) / 0.0596 < ( - _{} ) / _{} < (0.75 - 0.65) / 0.0596]
= P(-0.84 < z < 1.68)
= P(z < 1.68) - P(z < -0.84)
= 0.7531
Proportion = 0.7531
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