Questions: Suppose 52% of the population has a college degree. If a random sample of size 563 is selected, what is the probability that the proportion of persons with a college degree will differ from the population proportion by less than 5%?

Solution Steps

The standard error of the sample proportion is given by:

SE = sqrt[p(1-p)/n] = sqrt[0.52 * (1-0.52)/563] ≈ 0.0206

We want to find the probability that the sample proportion differs from the population proportion by less than 0.05. This can be written as:

P(0.52 - 0.05 < p̂ < 0.52 + 0.05) = P(0.47 < p̂ < 0.57)

Convert these bounds into z-scores:

z1 = (0.47 - 0.52)/0.0206 ≈ -2.43 z2 = (0.57 - 0.52)/0.0206 ≈ 2.43

We want to find P(-2.43 < z < 2.43). Using a standard normal table or calculator, we find:

P(z < 2.43) ≈ 0.9925 P(z < -2.43) ≈ 0.0075

So, P(-2.43 < z < 2.43) = P(z < 2.43) - P(z < -2.43) = 0.9925 - 0.0075 = 0.985

Final Answer: The probability is approximately 0.985 or 98.5%.