Answer:
Explanation:
Given:
Proportion of students who are late for class in the population: p = 0.12
Sample size: n = 538
Margin of error: E = 0.03
The standard error of the proportion can be calculated using the formula:
SE = sqrt(p*(1-p)/n)
SE = sqrt(0.12*0.88/538)
SE = 0.019
The margin of error corresponds to a z-score at the 97.5th percentile (since we want the probability of the sample proportion being within 3% of the population proportion in either direction, which gives us a two-tailed test with an alpha level of 0.025 on each side). This z-score can be found using a standard normal distribution table or calculator and is approximately 1.96.
The margin of error can also be calculated using the formula:
E = z*(SE)
0.03 = 1.96*(0.019)
To find the probability that the sample proportion differs from the population proportion by less than 3%, we need to find the area under the standard normal distribution curve between the z-scores of -1.96 and 1.96. This area is equivalent to the probability of the sample proportion being within 3% of the population proportion. This probability can be found using a standard normal distribution table or calculator and is approximately 0.8534.
Therefore, the probability that the proportion of late students in a sample of 538 students would differ from the population proportion by less than 3% is 0.8534 or 85.34% (rounded to four decimal places)