A population proportion is .75. A random sample of size 100 will be taken and the sample proportion will be used to estimate the population proportion. What is the probability that the sample proportion will be within +/- .03 of the population proportion?

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A population proportion is .75.  A random sample of size 100 will be taken and the sample proportion  will be used to estimate the population proportion.  What is the probability that the sample proportion will be within +/- .03 of the population proportion?

BlackTom AI · Accepted Answer

We start by identifying the sampling distribution of the sample proportion. With a population proportion p = 0.75 and sample size n = 100, the standard error of the sample proportion p̂ is SE = sqrt[p(1 − p)/n] = sqrt[(0.75)(0.25)/100] = sqrt(0.1875/100) = sqrt(0.001875) ≈ 0.0433.

We want the probability that p̂ is within ±0.03 of the population proportion, i.e., P(|p̂ − 0.75| ≤ 0.03).

Convert the interval to standard normal Z using Z = (p̂ − p)/SE. The endpoints become Z = ±0.03 / 0.0433 ≈ ±0.693.

Thus the probability is P(−0.693 ≤ Z ≤ 0.693) = Φ(0.693) − Φ(−0.693) = 2Φ(0.693) − 1.

Consulting the standard normal CDF (or using a Z-table), Φ(0.693) ≈ 0.755. Therefore, the probability is 2 × 0.755 − 1 ≈ 0.510, i.e., about 51%.

Option-by-option assessment:
- 51%: This value matches the normal-approximation calculation above, so it is consistent with the distributional result.
- 75%: This would imply a much larger interval coverage than the actual spread given by the SE; it ignores the precise z-score corresponding to ±0.03 and overestimates the probability.
- 24%: This would correspond to a much smaller window (or a much larger SE) and is not aligned with p̂’s distribution here; it underestimates the true probability.
- 68%: While 68% is familiar as one-standard-deviation coverage for a standard normal, the interval here corresponds to about ±0.693 standard deviations, not ±1 SD, so 68% is not the correct coverage for this specific z-range.

In summary, the calculation using the normal approximation yields a probability near 0.51, making the 51% option the one that aligns with the proper computation.

A population proportion is .75. A random sample of size 100 will be taken and the sample proportion will be used to estimate the population proportion. What is the probability that the sample proportion will be within +/- .03 of the population proportion?

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