A sample of 51 observations will be taken from an infinite population. The population proportion equals .85. The probability that the sample proportion will be between .9115 and .9645 is _____.

Question

A sample of 51 observations will be taken from an infinite population. The population proportion equals .85. The probability that the sample proportion will be between .9115 and .9645 is _____.

BlackTom AI · Accepted Answer

To tackle this problem, I start by recognizing that the sampling distribution of the sample proportion p̂ from a large (infinite) population is approximately normal with mean p and variance p(1−p)/n when n is reasonably large.

First, compute the mean and standard deviation using p = 0.85 and n = 51:
- Mean (expected value) of p̂: μ = p = 0.85
- Variance of p̂: Var(p̂) = p(1−p)/n = 0.85 × 0.15 / 51 = 0.1275 / 51 ≈ 0.0025
- Standard deviation of p̂: σ = sqrt(0.0025) = 0.05

Now translate the probability statement into z-scores for the normal approximation. We want P(0.9115 ≤ p̂ ≤ 0.9645).
- Lower z: z1 = (0.9115 − 0.85) / 0.05 = 0.0615 / 0.05 = 1.23
- Upper z: z2 = (0.9645 − 0.85) / 0.05 = 0.1145 / 0.05 = 2.29

Using the standard normal distribution, the probability that Z lies between 1.23 and 2.29 is P(1.23 ≤ Z ≤ 2.29) = Φ(2.29) − Φ(1.23).
- Φ(2.29) ≈ 0.9890
- Φ(1.23) ≈ 0.8907
- Difference ≈ 0.9890 − 0.8907 ≈ 0.0983

Thus, the probability is about 0.0983, which matches the option provided as .0983.

Now evaluate each answer option:
- Option ".0819": This value would correspond to a slightly smaller interval or a different standard deviation; given the calculations above, the actual probability is larger, around 0.0983, so this is not correct.
- Option ".0983": This aligns closely with the computed probability using the normal approximation with the given p, n, and bounds, making it the correct estimate.
- Option ".8633": This would imply an extremely high probability, nearly 86%, which would occur if the interval captured nearly all the distribution; our interval from z ≈ 1.23 to 2.29 covers only a middle portion, so this is incorrect.
- Option "None of the other three": Since option ".0983" is a plausible and supported estimate, this choice is incorrect.

In summary, by mapping the sample-proportion interval to z-scores and using the normal approximation, the probability is approximately 0.0983, and the other options do not align with that calculation.

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A sample of 51 observations will be taken from an infinite population. The population proportion equals .85. The probability that the sample proportion will be between .9115 and .9645 is _____.

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