Final answer:
Using the normal approximation to the binomial distribution, the probability that the sample proportion will differ from the true population proportion of 0.07 by less than 0.05 in a sample of 213 is approximately 0.9956.
Step-by-step explanation:
To estimate the probability that the sample proportion will differ from the true population proportion by less than 0.05 when the true proportion is 0.07 and the sample size is 213, we need to use the normal approximation to the binomial distribution (provided that the sample size is large enough and the np and n(1-p) conditions are satisfied).
The standard error (SE) of the sampling distribution of the proportion is given by:
SE = sqrt(p(1-p)/n), where p is the population proportion and n is the sample size.
For the given problem:
SE = sqrt(0.07 * 0.93 / 213) = sqrt(0.0651 / 213) = sqrt(0.0003056) ≈ 0.0175
We are looking for the probability that the sample proportion is within 0.05 of the population proportion, which means we are interested in the range from p - 0.05 to p + 0.05, i.e., from 0.02 to 0.12.
This will be a two-tailed test, so we use the z-score formula:
Z = (x - p) / SE
The z-scores for both the lower and upper limits are:
- Z_lower = (0.02 - 0.07) / 0.0175 ≈ -2.8571
- Z_upper = (0.12 - 0.07) / 0.0175 ≈ 2.8571
Using standard z-tables or a statistical calculator, we can find the probabilities associated with these z-scores and subtract the lower tail probability from the upper tail probability to find the probability that the sample proportion is within 0.05 of the true proportion.
Assuming a normal distribution, the probability for Z < 2.8571 is approximately 0.9977 and for Z < -2.8571 it is approximately 0.0021. Thus, the probability that the sample proportion differs from the true population proportion by less than 0.05 is approximately:
0.9977 - 0.0021 = 0.9956.
The answer rounded to four decimal places would be 0.9956.