Final answer:
To find the probability that more than 73% out of 300 of the sample have a driver's license, we need to use the normal distribution. We can calculate the z-score for the given percentage and then find the corresponding probability in the standard normal distribution table. The probability of more than 73% will be 1 minus the probability found in the table.
Step-by-step explanation:
To find the probability that more than 73% out of 300 of the sample have a driver's license, we need to use the normal distribution. Assuming that the distribution of driver's licenses in the sample follows a normal distribution, we can calculate the z-score for the given percentage.
First, let's find the mean and standard deviation of the sample. The mean is 73% of 300, which is 219. The standard deviation is the square root of the product of the sample size, the probability of success (1 - 0.73), and the probability of failure (0.73).
Once we have the mean and standard deviation, we can calculate the z-score using the formula: z = (x - mean) / standard deviation, where x is the desired percentage (73%) converted to a decimal.
Finally, we can look up the corresponding probability in the standard normal distribution table using the z-score. The probability of more than 73% out of 300 of the sample having a driver's license will be 1 minus the probability found in the table. Therefore, the correct answer is d) 1.0.