Use a normal approximation to find the probability of the indicated number of voters. In this case assume that 119 eligible voters aged 18-24 are randomly selected. Suppose a previous study showed that among eligible voters aged 18-24, 22% of them voted.

Question

The probability that fewer than 32 of 19 eligible voters voted is=

Answer 1

To find the probability using a normal approximation, we'll first calculate the mean and standard deviation.

Mean (μ) = np = 119 * 0.22 = 26.18
Standard deviation (σ) = √(np(1-p)) = √(119 * 0.22 * 0.78) = √(21.5884) ≈ 4.65

Now, we will find the z-score for 32 voters.

z = (x - μ) / σ = (32 - 26.18) / 4.65 ≈ 1.25

Using a z-table, we find the probability for a z-score of 1.25 is 0.8944. However, we want the probability that fewer than 32 voted, so we'll find the complement.

P(x < 32) = 1 - P(x >= 32) = 1 - 0.8944 = 0.1056

Thus, the probability that fewer than 32 of the 119 eligible voters aged 18-24 voted is approximately 0.1056 or 10.56%.