Pew Research says only 40% of Americans plan to get a COVID vaccine. If this is the true population proportion, and a random sample of 100 are asked if they will get a COVID vaccine, what is the probability of 50% or more saying they will?

Question

Pew Research says only 40% of Americans plan to get a COVID vaccine.  If this is the true population proportion, and a random sample of 100 are asked if they will get a COVID vaccine, what is the probability of 50% or more saying they will?

BlackTom AI · Accepted Answer

The question gives a population proportion p = 0.40 and a sample size n = 100. We want the probability that 50% or more of the 100 sampled individuals say they will get a COVID vaccine, i.e., that X ≥ 50 where X ~ Binomial(n = 100, p = 0.40).

Option 1: 43%
- This would imply a probability around 0.43, which is far too large given the mean of the distribution is np = 40 and 50 is more than one standard deviation above the mean. The standard deviation is sqrt(np(1−p)) = sqrt(100 * 0.4 * 0.6) = sqrt(24) ≈ 4.90. The value 50 is (50 − 40) / 4.90 ≈ 2.04 standard deviations above the mean, so the tail probability should be well below 0.5, not near 0.43. This option overstates the likelihood considerably.

Option 2: 60%
- An outcome with probability 0.60 would imply a majority of samples yield 50 or more yes responses in a way that is much more common than the binomial distribution would suggest. Since the mean is 40 and 50 is far into the upper tail (roughly two standard deviations away), the probability cannot be this large. This option contradicts the distribution’s typical tail behavior.

Option 3: 19%
- This is closer to the ballpark for a tail probability, but still a bit high for P(X ≥ 50) given the mean is 40 and the standard deviation is about 4.90. The z-score for 49.5 (with continuity correction for a discrete binomial when approximating with a normal) is (49.5 − 40) / 4.90 ≈ 1.95, which corresponds to a tail probability around 0.025 when using standard normal tables. Even without continuity correction, the approximate z for 50 is (50 − 40) / 4.90 ≈ 2.04, which yields a tail probability around 0.020. Therefore 19% is too large for this tail probability.

Option 4: 2%
- This is the most plausible among the options. Using either a normal approximation with continuity correction or a binomial tail calculation, P(X ≥ 50) is about 0.02–0.03. Specifically, the z-score with continuity correction is (49.5 − 40) / 4.90 ≈ 1.95, giving a tail probability near 0.025. Without correction, the z-score is about 2.04, giving a tail around 0.020. Both routes place the probability in the few-percent range, roughly 2–3%, which aligns with this option.

Additional notes:
- If you choose to compute via the sample proportion p̂ = X/100, then p̂ has approximately a normal distribution with mean p = 0.40 and standard error sqrt(p(1−p)/n) = sqrt(0.24/100) ≈ 0.049. The target p̂ ≥ 0.50 corresponds to a z = (0.50 − 0.40) / 0.049 ≈ 2.04, again yielding a tail probability around 0.02.
- The continuity correction helps to better approximate the discrete binomial with a continuous normal, nudging the estimate slightly to reflect the discrete gap between X = 49 and X = 50.
- Across these calculations, the consistent story is that getting 50 or more affirmative responses in a sample of 100 when p = 0.40 is a relatively unlikely event, on the order of a few percent, not near half or a fifth of the time.

Pew Research says only 40% of Americans plan to get a COVID vaccine. If this is the true population proportion, and a random sample of 100 are asked if they will get a COVID vaccine, what is the probability of 50% or more saying they will?

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