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

We begin by restating the problem to ensure clarity: a sample of 51 observations is drawn from an infinite population with population proportion p = 0.85. We want the probability that the sample proportion p̂ falls between 0.9115 and 0.9645. The standard approach is to treat p̂ as approximately normal with mean μ = p and standard deviation σ = sqrt[p(1−p)/n]. Here, p = 0.85 and n = 51.

First, compute the standard deviation: p(1−p) = 0.85 × 0.15 = 0.1275. Divide by n: 0.1275 / 51 = 0.0025. The standard deviation is the square root of this variance, so σ = sqrt(0.0025) = 0.05.

Next, convert the interval for p̂ into z-scores using z = (p̂ − μ) / σ.
- For the left endpoint p̂ = 0.9115: z1 = (0.9115 − 0.85) / 0.05 = 0.0615 / 0.05 = 1.23.
- For the right endpoint p̂ = 0.9645: z2 = (0.9645 − 0.85) / 0.05 = 0.1145 / 0.05 = 2.29.

Now we look up or approximate the standard normal probabilities between z = 1.23 and z = 2.29. Using standard normal CDF values (or a calculator):
- Φ(2.29) is approximately 0.989, or 0.9890 to four decimals.
- Φ(1.23) is approximately 0.891, or 0.8910 to four decimals.

The probability that z lies between 1.23 and 2.29 is Φ(2.29) − Φ(1.23) ≈ 0.9890 − 0.8910 = 0.0980, which is very close to 0.0983 when using more precise values.

Option-by-option analysis:
- Option ".0983": This matches the calculated probability using the normal approximation, given the mean, variance, and z-score bounds. The difference between 0.0980 and 0.0983 arises from rounding in the CDF values, but 0.0983 is consistent with the normal-approximation result.
- Option "None of the other three": This would only be correct if none of the listed numbers matched the computed probability. Since 0.0983 aligns with the calculation, this option is not appropriate.
- Option ".8633": This value is far from the calculated probability. It would imply a probability around 86%, which would require a much wider interval or a much smaller standard deviation; neither is the case here.
- Option ".0819": This is a plausible alternative if using cruder approximations or slightly different z-values, but it does not align as closely with the detailed calculation that yields about 0.0983. The discrepancy suggests either rounding differences or a different method, but the direct normal-approximation calculation supports the 0.0983 figure.

In summary, the central steps are identifying the sampling distribution of p̂ as approximately normal with mean p and sd sqrt[p(1−p)/n], converting the p̂ interval to z-scores, and evaluating the normal CDF difference between those z-values. The resulting probability is about 0.0983, making the first option the best match to the computed value, while the other options either overstate, understate, or misstate the likelihood based on the interval and variance involved.

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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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